How to give the string during runtime rather than pre initialization to a character pointer say char *b; through keyboard?
First, to clear things up a bit, as per C11 standard, chapter §7.1.1,
A string is a contiguous sequence of characters terminated by and including the first null
character.
and, as per §6.4.5
A character string literal is a sequence of zero or more multibyte characters enclosed in
double-quotes, as in "xyz".
So, they are not the same.
However, to input a string from user, you can follow either of the below cases
define a char array and scan the input (scanf(), fgets()).
define a pointer, allocate memory and then use scanf() or fgets() to read the input from the user.
You can define your string before compile and running your code with following syntax:
char * str = "Hello World";
by this way you define a constant string that change its content result in unspecific behavior. if you want you can allocate a string pointer and then use functions like scanf(), fgets() and ... in order to get its content from user or you can use functions like sprintf() in order to fill your string in your program. for example:
char *str = malloc(sizeof(char) * 20);
sprintf(str, "%s", Hello World");
If you want to have dynamic size string in runtime you can implement something like C++ vectors
and store your string in it.
Related
I understand that strings in C are just character arrays. So I tried the following code, but it gives strange results, such as garbage output or program crashes:
#include <stdio.h>
int main (void)
{
char str [5] = "hello";
puts(str);
}
Why doesn't this work?
It compiles cleanly with gcc -std=c17 -pedantic-errors -Wall -Wextra.
Note: This post is meant to be used as a canonical FAQ for problems stemming from a failure to allocate room for a NUL terminator when declaring a string.
A C string is a character array that ends with a null terminator.
All characters have a symbol table value. The null terminator is the symbol value 0 (zero). It is used to mark the end of a string. This is necessary since the size of the string isn't stored anywhere.
Therefore, every time you allocate room for a string, you must include sufficient space for the null terminator character. Your example does not do this, it only allocates room for the 5 characters of "hello". Correct code should be:
char str[6] = "hello";
Or equivalently, you can write self-documenting code for 5 characters plus 1 null terminator:
char str[5+1] = "hello";
But you can also use this and let the compiler do the counting and pick the size:
char str[] = "hello"; // Will allocate 6 bytes automatically
When allocating memory for a string dynamically in run-time, you also need to allocate room for the null terminator:
char input[n] = ... ;
...
char* str = malloc(strlen(input) + 1);
If you don't append a null terminator at the end of a string, then library functions expecting a string won't work properly and you will get "undefined behavior" bugs such as garbage output or program crashes.
The most common way to write a null terminator character in C is by using a so-called "octal escape sequence", looking like this: '\0'. This is 100% equivalent to writing 0, but the \ serves as self-documenting code to state that the zero is explicitly meant to be a null terminator. Code such as if(str[i] == '\0') will check if the specific character is the null terminator.
Please note that the term null terminator has nothing to do with null pointers or the NULL macro! This can be confusing - very similar names but very different meanings. This is why the null terminator is sometimes referred to as NUL with one L, not to be confused with NULL or null pointers. See answers to this SO question for further details.
The "hello" in your code is called a string literal. This is to be regarded as a read-only string. The "" syntax means that the compiler will append a null terminator in the end of the string literal automatically. So if you print out sizeof("hello") you will get 6, not 5, because you get the size of the array including a null terminator.
It compiles cleanly with gcc
Indeed, not even a warning. This is because of a subtle detail/flaw in the C language that allows character arrays to be initialized with a string literal that contains exactly as many characters as there is room in the array and then silently discard the null terminator (C17 6.7.9/15). The language is purposely behaving like this for historical reasons, see Inconsistent gcc diagnostic for string initialization for details. Also note that C++ is different here and does not allow this trick/flaw to be used.
From the C Standard (7.1.1 Definitions of terms)
1 A string is a contiguous sequence of characters terminated by and
including the first null character. The term multibyte string is
sometimes used instead to emphasize special processing given to
multibyte characters contained in the string or to avoid confusion
with a wide string. A pointer to a string is a pointer to its initial
(lowest addressed) character. The length of a string is the number of
bytes preceding the null character and the value of a string is the
sequence of the values of the contained characters, in order.
In this declaration
char str [5] = "hello";
the string literal "hello" has the internal representation like
{ 'h', 'e', 'l', 'l', 'o', '\0' }
so it has 6 characters including the terminating zero. Its elements are used to initialize the character array str which reserve space only for 5 characters.
The C Standard (opposite to the C++ Standard) allows such an initialization of a character array when the terminating zero of a string literal is not used as an initializer.
However as a result the character array str does not contain a string.
If you want that the array would contain a string you could write
char str [6] = "hello";
or just
char str [] = "hello";
In the last case the size of the character array is determined from the number of initializers of the string literal that is equal to 6.
Can all strings be considered an array of characters (Yes), can all character arrays be considered strings (No).
Why Not? and Why does it matter?
In addition to the other answers explaining that the length of a string is not stored anywhere as part of the string and the references to the standard where a string is defined, the flip-side is "How do the C library functions handle strings?"
While a character array can hold the same characters, it is simply an array of characters unless the last character is followed by the nul-terminating character. That nul-terminating character is what allows the array of characters to be considered (handled as) a string.
All functions in C that expect a string as an argument expect the sequence of characters to be nul-terminated. Why?
It has to do with the way all string functions work. Since the length isn't included as part of an array, string-functions, scan forward in the array until the nul-character (e.g. '\0' -- equivalent to decimal 0) is found. See ASCII Table and Description. Regardless whether you are using strcpy, strchr, strcspn, etc.. All string functions rely on the nul-terminating character being present to define where the end of that string is.
A comparison of two similar functions from string.h will emphasize the importance of the nul-terminating character. Take for example:
char *strcpy(char *dest, const char *src);
The strcpy function simply copies bytes from src to dest until the nul-terminating character is found telling strcpy where to stop copying characters. Now take the similar function memcpy:
void *memcpy(void *dest, const void *src, size_t n);
The function performs a similar operation, but does not consider or require the src parameter to be a string. Since memcpy cannot simply scan forward in src copying bytes to dest until a nul-terminating character is reached, it requires an explicit number of bytes to copy as a third parameter. This third parameter provides memcpy with the same size information strcpy is able to derive simply by scanning forward until a nul-terminating character is found.
(which also emphasizes what goes wrong in strcpy (or any function expecting a string) if you fail to provide the function with a nul-terminated string -- it has no idea where to stop and will happily race off across the rest of your memory segment invoking Undefined Behavior until a nul-character just happens to be found somewhere in memory -- or a Segmentation Fault occurs)
That is why functions expecting a nul-terminated string must be passed a nul-terminated string and why it matters.
Intuitively...
Think of an array as a variable (holds things) and a string as a value (can be placed in a variable).
They are certainly not the same thing. In your case the variable is too small to hold the string, so the string gets cut off. ("quoted strings" in C have an implicit null character at the end.)
However it's possible to store a string in an array that is much larger than the string.
Note that the usual assignment and comparison operators (= == < etc.) don't work as you might expect. But the strxyz family of functions comes pretty close, once you know what you're doing. See the C FAQ on strings and arrays.
I understand that strings in C are just character arrays. So I tried the following code, but it gives strange results, such as garbage output or program crashes:
#include <stdio.h>
int main (void)
{
char str [5] = "hello";
puts(str);
}
Why doesn't this work?
It compiles cleanly with gcc -std=c17 -pedantic-errors -Wall -Wextra.
Note: This post is meant to be used as a canonical FAQ for problems stemming from a failure to allocate room for a NUL terminator when declaring a string.
A C string is a character array that ends with a null terminator.
All characters have a symbol table value. The null terminator is the symbol value 0 (zero). It is used to mark the end of a string. This is necessary since the size of the string isn't stored anywhere.
Therefore, every time you allocate room for a string, you must include sufficient space for the null terminator character. Your example does not do this, it only allocates room for the 5 characters of "hello". Correct code should be:
char str[6] = "hello";
Or equivalently, you can write self-documenting code for 5 characters plus 1 null terminator:
char str[5+1] = "hello";
But you can also use this and let the compiler do the counting and pick the size:
char str[] = "hello"; // Will allocate 6 bytes automatically
When allocating memory for a string dynamically in run-time, you also need to allocate room for the null terminator:
char input[n] = ... ;
...
char* str = malloc(strlen(input) + 1);
If you don't append a null terminator at the end of a string, then library functions expecting a string won't work properly and you will get "undefined behavior" bugs such as garbage output or program crashes.
The most common way to write a null terminator character in C is by using a so-called "octal escape sequence", looking like this: '\0'. This is 100% equivalent to writing 0, but the \ serves as self-documenting code to state that the zero is explicitly meant to be a null terminator. Code such as if(str[i] == '\0') will check if the specific character is the null terminator.
Please note that the term null terminator has nothing to do with null pointers or the NULL macro! This can be confusing - very similar names but very different meanings. This is why the null terminator is sometimes referred to as NUL with one L, not to be confused with NULL or null pointers. See answers to this SO question for further details.
The "hello" in your code is called a string literal. This is to be regarded as a read-only string. The "" syntax means that the compiler will append a null terminator in the end of the string literal automatically. So if you print out sizeof("hello") you will get 6, not 5, because you get the size of the array including a null terminator.
It compiles cleanly with gcc
Indeed, not even a warning. This is because of a subtle detail/flaw in the C language that allows character arrays to be initialized with a string literal that contains exactly as many characters as there is room in the array and then silently discard the null terminator (C17 6.7.9/15). The language is purposely behaving like this for historical reasons, see Inconsistent gcc diagnostic for string initialization for details. Also note that C++ is different here and does not allow this trick/flaw to be used.
From the C Standard (7.1.1 Definitions of terms)
1 A string is a contiguous sequence of characters terminated by and
including the first null character. The term multibyte string is
sometimes used instead to emphasize special processing given to
multibyte characters contained in the string or to avoid confusion
with a wide string. A pointer to a string is a pointer to its initial
(lowest addressed) character. The length of a string is the number of
bytes preceding the null character and the value of a string is the
sequence of the values of the contained characters, in order.
In this declaration
char str [5] = "hello";
the string literal "hello" has the internal representation like
{ 'h', 'e', 'l', 'l', 'o', '\0' }
so it has 6 characters including the terminating zero. Its elements are used to initialize the character array str which reserve space only for 5 characters.
The C Standard (opposite to the C++ Standard) allows such an initialization of a character array when the terminating zero of a string literal is not used as an initializer.
However as a result the character array str does not contain a string.
If you want that the array would contain a string you could write
char str [6] = "hello";
or just
char str [] = "hello";
In the last case the size of the character array is determined from the number of initializers of the string literal that is equal to 6.
Can all strings be considered an array of characters (Yes), can all character arrays be considered strings (No).
Why Not? and Why does it matter?
In addition to the other answers explaining that the length of a string is not stored anywhere as part of the string and the references to the standard where a string is defined, the flip-side is "How do the C library functions handle strings?"
While a character array can hold the same characters, it is simply an array of characters unless the last character is followed by the nul-terminating character. That nul-terminating character is what allows the array of characters to be considered (handled as) a string.
All functions in C that expect a string as an argument expect the sequence of characters to be nul-terminated. Why?
It has to do with the way all string functions work. Since the length isn't included as part of an array, string-functions, scan forward in the array until the nul-character (e.g. '\0' -- equivalent to decimal 0) is found. See ASCII Table and Description. Regardless whether you are using strcpy, strchr, strcspn, etc.. All string functions rely on the nul-terminating character being present to define where the end of that string is.
A comparison of two similar functions from string.h will emphasize the importance of the nul-terminating character. Take for example:
char *strcpy(char *dest, const char *src);
The strcpy function simply copies bytes from src to dest until the nul-terminating character is found telling strcpy where to stop copying characters. Now take the similar function memcpy:
void *memcpy(void *dest, const void *src, size_t n);
The function performs a similar operation, but does not consider or require the src parameter to be a string. Since memcpy cannot simply scan forward in src copying bytes to dest until a nul-terminating character is reached, it requires an explicit number of bytes to copy as a third parameter. This third parameter provides memcpy with the same size information strcpy is able to derive simply by scanning forward until a nul-terminating character is found.
(which also emphasizes what goes wrong in strcpy (or any function expecting a string) if you fail to provide the function with a nul-terminated string -- it has no idea where to stop and will happily race off across the rest of your memory segment invoking Undefined Behavior until a nul-character just happens to be found somewhere in memory -- or a Segmentation Fault occurs)
That is why functions expecting a nul-terminated string must be passed a nul-terminated string and why it matters.
Intuitively...
Think of an array as a variable (holds things) and a string as a value (can be placed in a variable).
They are certainly not the same thing. In your case the variable is too small to hold the string, so the string gets cut off. ("quoted strings" in C have an implicit null character at the end.)
However it's possible to store a string in an array that is much larger than the string.
Note that the usual assignment and comparison operators (= == < etc.) don't work as you might expect. But the strxyz family of functions comes pretty close, once you know what you're doing. See the C FAQ on strings and arrays.
I understand that strings in C are just character arrays. So I tried the following code, but it gives strange results, such as garbage output or program crashes:
#include <stdio.h>
int main (void)
{
char str [5] = "hello";
puts(str);
}
Why doesn't this work?
It compiles cleanly with gcc -std=c17 -pedantic-errors -Wall -Wextra.
Note: This post is meant to be used as a canonical FAQ for problems stemming from a failure to allocate room for a NUL terminator when declaring a string.
A C string is a character array that ends with a null terminator.
All characters have a symbol table value. The null terminator is the symbol value 0 (zero). It is used to mark the end of a string. This is necessary since the size of the string isn't stored anywhere.
Therefore, every time you allocate room for a string, you must include sufficient space for the null terminator character. Your example does not do this, it only allocates room for the 5 characters of "hello". Correct code should be:
char str[6] = "hello";
Or equivalently, you can write self-documenting code for 5 characters plus 1 null terminator:
char str[5+1] = "hello";
But you can also use this and let the compiler do the counting and pick the size:
char str[] = "hello"; // Will allocate 6 bytes automatically
When allocating memory for a string dynamically in run-time, you also need to allocate room for the null terminator:
char input[n] = ... ;
...
char* str = malloc(strlen(input) + 1);
If you don't append a null terminator at the end of a string, then library functions expecting a string won't work properly and you will get "undefined behavior" bugs such as garbage output or program crashes.
The most common way to write a null terminator character in C is by using a so-called "octal escape sequence", looking like this: '\0'. This is 100% equivalent to writing 0, but the \ serves as self-documenting code to state that the zero is explicitly meant to be a null terminator. Code such as if(str[i] == '\0') will check if the specific character is the null terminator.
Please note that the term null terminator has nothing to do with null pointers or the NULL macro! This can be confusing - very similar names but very different meanings. This is why the null terminator is sometimes referred to as NUL with one L, not to be confused with NULL or null pointers. See answers to this SO question for further details.
The "hello" in your code is called a string literal. This is to be regarded as a read-only string. The "" syntax means that the compiler will append a null terminator in the end of the string literal automatically. So if you print out sizeof("hello") you will get 6, not 5, because you get the size of the array including a null terminator.
It compiles cleanly with gcc
Indeed, not even a warning. This is because of a subtle detail/flaw in the C language that allows character arrays to be initialized with a string literal that contains exactly as many characters as there is room in the array and then silently discard the null terminator (C17 6.7.9/15). The language is purposely behaving like this for historical reasons, see Inconsistent gcc diagnostic for string initialization for details. Also note that C++ is different here and does not allow this trick/flaw to be used.
From the C Standard (7.1.1 Definitions of terms)
1 A string is a contiguous sequence of characters terminated by and
including the first null character. The term multibyte string is
sometimes used instead to emphasize special processing given to
multibyte characters contained in the string or to avoid confusion
with a wide string. A pointer to a string is a pointer to its initial
(lowest addressed) character. The length of a string is the number of
bytes preceding the null character and the value of a string is the
sequence of the values of the contained characters, in order.
In this declaration
char str [5] = "hello";
the string literal "hello" has the internal representation like
{ 'h', 'e', 'l', 'l', 'o', '\0' }
so it has 6 characters including the terminating zero. Its elements are used to initialize the character array str which reserve space only for 5 characters.
The C Standard (opposite to the C++ Standard) allows such an initialization of a character array when the terminating zero of a string literal is not used as an initializer.
However as a result the character array str does not contain a string.
If you want that the array would contain a string you could write
char str [6] = "hello";
or just
char str [] = "hello";
In the last case the size of the character array is determined from the number of initializers of the string literal that is equal to 6.
Can all strings be considered an array of characters (Yes), can all character arrays be considered strings (No).
Why Not? and Why does it matter?
In addition to the other answers explaining that the length of a string is not stored anywhere as part of the string and the references to the standard where a string is defined, the flip-side is "How do the C library functions handle strings?"
While a character array can hold the same characters, it is simply an array of characters unless the last character is followed by the nul-terminating character. That nul-terminating character is what allows the array of characters to be considered (handled as) a string.
All functions in C that expect a string as an argument expect the sequence of characters to be nul-terminated. Why?
It has to do with the way all string functions work. Since the length isn't included as part of an array, string-functions, scan forward in the array until the nul-character (e.g. '\0' -- equivalent to decimal 0) is found. See ASCII Table and Description. Regardless whether you are using strcpy, strchr, strcspn, etc.. All string functions rely on the nul-terminating character being present to define where the end of that string is.
A comparison of two similar functions from string.h will emphasize the importance of the nul-terminating character. Take for example:
char *strcpy(char *dest, const char *src);
The strcpy function simply copies bytes from src to dest until the nul-terminating character is found telling strcpy where to stop copying characters. Now take the similar function memcpy:
void *memcpy(void *dest, const void *src, size_t n);
The function performs a similar operation, but does not consider or require the src parameter to be a string. Since memcpy cannot simply scan forward in src copying bytes to dest until a nul-terminating character is reached, it requires an explicit number of bytes to copy as a third parameter. This third parameter provides memcpy with the same size information strcpy is able to derive simply by scanning forward until a nul-terminating character is found.
(which also emphasizes what goes wrong in strcpy (or any function expecting a string) if you fail to provide the function with a nul-terminated string -- it has no idea where to stop and will happily race off across the rest of your memory segment invoking Undefined Behavior until a nul-character just happens to be found somewhere in memory -- or a Segmentation Fault occurs)
That is why functions expecting a nul-terminated string must be passed a nul-terminated string and why it matters.
Intuitively...
Think of an array as a variable (holds things) and a string as a value (can be placed in a variable).
They are certainly not the same thing. In your case the variable is too small to hold the string, so the string gets cut off. ("quoted strings" in C have an implicit null character at the end.)
However it's possible to store a string in an array that is much larger than the string.
Note that the usual assignment and comparison operators (= == < etc.) don't work as you might expect. But the strxyz family of functions comes pretty close, once you know what you're doing. See the C FAQ on strings and arrays.
I understand that strings in C are just character arrays. So I tried the following code, but it gives strange results, such as garbage output or program crashes:
#include <stdio.h>
int main (void)
{
char str [5] = "hello";
puts(str);
}
Why doesn't this work?
It compiles cleanly with gcc -std=c17 -pedantic-errors -Wall -Wextra.
Note: This post is meant to be used as a canonical FAQ for problems stemming from a failure to allocate room for a NUL terminator when declaring a string.
A C string is a character array that ends with a null terminator.
All characters have a symbol table value. The null terminator is the symbol value 0 (zero). It is used to mark the end of a string. This is necessary since the size of the string isn't stored anywhere.
Therefore, every time you allocate room for a string, you must include sufficient space for the null terminator character. Your example does not do this, it only allocates room for the 5 characters of "hello". Correct code should be:
char str[6] = "hello";
Or equivalently, you can write self-documenting code for 5 characters plus 1 null terminator:
char str[5+1] = "hello";
But you can also use this and let the compiler do the counting and pick the size:
char str[] = "hello"; // Will allocate 6 bytes automatically
When allocating memory for a string dynamically in run-time, you also need to allocate room for the null terminator:
char input[n] = ... ;
...
char* str = malloc(strlen(input) + 1);
If you don't append a null terminator at the end of a string, then library functions expecting a string won't work properly and you will get "undefined behavior" bugs such as garbage output or program crashes.
The most common way to write a null terminator character in C is by using a so-called "octal escape sequence", looking like this: '\0'. This is 100% equivalent to writing 0, but the \ serves as self-documenting code to state that the zero is explicitly meant to be a null terminator. Code such as if(str[i] == '\0') will check if the specific character is the null terminator.
Please note that the term null terminator has nothing to do with null pointers or the NULL macro! This can be confusing - very similar names but very different meanings. This is why the null terminator is sometimes referred to as NUL with one L, not to be confused with NULL or null pointers. See answers to this SO question for further details.
The "hello" in your code is called a string literal. This is to be regarded as a read-only string. The "" syntax means that the compiler will append a null terminator in the end of the string literal automatically. So if you print out sizeof("hello") you will get 6, not 5, because you get the size of the array including a null terminator.
It compiles cleanly with gcc
Indeed, not even a warning. This is because of a subtle detail/flaw in the C language that allows character arrays to be initialized with a string literal that contains exactly as many characters as there is room in the array and then silently discard the null terminator (C17 6.7.9/15). The language is purposely behaving like this for historical reasons, see Inconsistent gcc diagnostic for string initialization for details. Also note that C++ is different here and does not allow this trick/flaw to be used.
From the C Standard (7.1.1 Definitions of terms)
1 A string is a contiguous sequence of characters terminated by and
including the first null character. The term multibyte string is
sometimes used instead to emphasize special processing given to
multibyte characters contained in the string or to avoid confusion
with a wide string. A pointer to a string is a pointer to its initial
(lowest addressed) character. The length of a string is the number of
bytes preceding the null character and the value of a string is the
sequence of the values of the contained characters, in order.
In this declaration
char str [5] = "hello";
the string literal "hello" has the internal representation like
{ 'h', 'e', 'l', 'l', 'o', '\0' }
so it has 6 characters including the terminating zero. Its elements are used to initialize the character array str which reserve space only for 5 characters.
The C Standard (opposite to the C++ Standard) allows such an initialization of a character array when the terminating zero of a string literal is not used as an initializer.
However as a result the character array str does not contain a string.
If you want that the array would contain a string you could write
char str [6] = "hello";
or just
char str [] = "hello";
In the last case the size of the character array is determined from the number of initializers of the string literal that is equal to 6.
Can all strings be considered an array of characters (Yes), can all character arrays be considered strings (No).
Why Not? and Why does it matter?
In addition to the other answers explaining that the length of a string is not stored anywhere as part of the string and the references to the standard where a string is defined, the flip-side is "How do the C library functions handle strings?"
While a character array can hold the same characters, it is simply an array of characters unless the last character is followed by the nul-terminating character. That nul-terminating character is what allows the array of characters to be considered (handled as) a string.
All functions in C that expect a string as an argument expect the sequence of characters to be nul-terminated. Why?
It has to do with the way all string functions work. Since the length isn't included as part of an array, string-functions, scan forward in the array until the nul-character (e.g. '\0' -- equivalent to decimal 0) is found. See ASCII Table and Description. Regardless whether you are using strcpy, strchr, strcspn, etc.. All string functions rely on the nul-terminating character being present to define where the end of that string is.
A comparison of two similar functions from string.h will emphasize the importance of the nul-terminating character. Take for example:
char *strcpy(char *dest, const char *src);
The strcpy function simply copies bytes from src to dest until the nul-terminating character is found telling strcpy where to stop copying characters. Now take the similar function memcpy:
void *memcpy(void *dest, const void *src, size_t n);
The function performs a similar operation, but does not consider or require the src parameter to be a string. Since memcpy cannot simply scan forward in src copying bytes to dest until a nul-terminating character is reached, it requires an explicit number of bytes to copy as a third parameter. This third parameter provides memcpy with the same size information strcpy is able to derive simply by scanning forward until a nul-terminating character is found.
(which also emphasizes what goes wrong in strcpy (or any function expecting a string) if you fail to provide the function with a nul-terminated string -- it has no idea where to stop and will happily race off across the rest of your memory segment invoking Undefined Behavior until a nul-character just happens to be found somewhere in memory -- or a Segmentation Fault occurs)
That is why functions expecting a nul-terminated string must be passed a nul-terminated string and why it matters.
Intuitively...
Think of an array as a variable (holds things) and a string as a value (can be placed in a variable).
They are certainly not the same thing. In your case the variable is too small to hold the string, so the string gets cut off. ("quoted strings" in C have an implicit null character at the end.)
However it's possible to store a string in an array that is much larger than the string.
Note that the usual assignment and comparison operators (= == < etc.) don't work as you might expect. But the strxyz family of functions comes pretty close, once you know what you're doing. See the C FAQ on strings and arrays.
how do i Initialize my code if all im using are words and no numbers?
I have been trying to just use char * but it is saying that its still not initialized
char *Carson;
printf("Enter a name:\n");
scanf("%s",Name);
printf("%s Hello Carson\n", Carson);
You either have to allocate memory dynamically and assign it to Carson (see e.g. `malloc? ), or make it an array. There's no way around it. And for that, the code must contain a number. The number could be input from the user though, so you won't have any actual numbers in the source.
Remember that in C all strings need an extra terminator character (added automatically by scanf) so remember to add space for it.
A solution without any number, I don't think this must be used for practical applications, just a hack
char Carson[sizeof(long long) * sizeof(long long)];
printf("Size = %d\n", sizeof Carson);
printf("Enter a name:\n");
scanf("%s",Carson);
printf("%s Hello Carson\n", Carson);
In my system it create a char array of 64 bytes = 8 * 8, the size of long long in most systems is 8 bytes although it's size depends on your compiler and operating system
you might like to initialize Carson like this:
char *Carson = malloc(sizeof(char)*200);/* for 200 characters */
Don't forget to add \0 terminator and also, donot forget to free it once you are done using it.
In order to initialize variables in C you need to use constants values, that is, expressions whose value can be known at compile time.
For integer or float types you can use mathematical formulas involving only constant operands, thus you can obtain still a constant value that can be used in a initiaiization.
What you call "words" have been called better "strings".
In C you are able to use strings that are constant at compile time, also called "string literals".
A string literal has to be indicated surrounded by quotes, like these examples:
"Hello world!"
"Peter & John"
"user#gmail.com"
and so on.
There are some rules that you need to remember: some special characters have "escape sequences" to be used inside a string literal.
Now you can use that string literals in order to initialize a char* variable:
char *name = "Mr. Smith";
char *city = "Amsterdam";
The result of the initialization gives a C string style, that is, an array of char object, whose length is the amount of quoted characters in the string literal, plus 1, because a null character is added at the end. Thus, in memory you have:
char *city ----> |A|m|s|t|e|r|d|a|m|\0|
Thus, city points to an array of 10 chars.
The last character, \0, means "null character", whose ASCII code is 0. Since it corresponds to a non-printable character, it has to be indicated with the escape sequence \0.
For more information, take a look on these websites:
Escape sequences in C
Storage of string literals
If you initialize a pointer to char object to a string literal, the compiler reserves memory automatically for you, son you don't need any malloc() at all.
However, you cannot modify the characters of such a string.
If you are interested in modify the characters, you can use better un array of char object:
char name[30] = "Schwarzenegger";
The array reserves 30 chars for the string literal "Schwarzenegger".
Only the first 14 are used for the string, plus 1 holding the null character attached at the end.
The rest of chars of the array have dummy information, but there is no problem because they are not printed. (The standard library functions always stop processing the string when they encounter a null character).
EDITED More information.
About your particular error message: "lack of initialization", the problem is that in the definition of the pointer to char object:
char *name;
you only have a "pointer to" an undefined block of memory.
You have to specify the array of char that name will be point to.
If you initialize with a string literal, there is not any problem, because the address of the string literal is passed to name.
But, since you are planning to use name for data input by means of scanf(), you have to allocate memory enough. You can do that other users have explained yet in their answers, that is, by using malloc().
I think there is need to do changes in your code,
char Carson = NULL;
Carson = (char)malloc(sizeof(char)*256);
printf("Enter a name:\n");
scanf("%s",Carson );
printf("%s Hello Carson\n", Carson);
in place of 256 u can use whatever value you want.
let me know if it works.