I want to copy all the characters in a char[] to another char[]. However, suppose there are is a '\0' then I want to treat that as a normal character--a literal if you will.
Therefore, when I printout the char[] with the format specifier %s, it should not stop in the middle.
e.g.
// chars copied to array x
char x[] = {'h','e','\0','l','l','o','\0'}
printf("%s\n",x); // prints 'he\0llo'
Is there a way to do this?
Use memcpy if you know how many characters you need to copy. If x is a real array as in your example, sizeof(x) will give you that amount, but if you pass x as a parameter to a function, sizeof will not work inside that function (it will just show the size of the pointer), so the basic rule is that your strings either should be NUL-terminated, or you should keep their size in a separate variable.
For printing, you can either print in the for loop as #alk suggests, or use fwrite to write any buffer (possibly with NULs) to stdout:
fwrite(x, sizeof(x), 1, stdout); /* sizeof will work only for a real array */
Wrong use of "%s". There is not a way to use printf("%s" to print an array with data after the first null character or '\0'.
char x[] = {'h','e','\0','l','l','o','\0'}
printf("%s\n",x); // This only prints "he"
printf("%s",... is for printing strings. x is a string only up to and including the first '\0'.
To "copy all the characters in a char[] to another char[].", use memcpy().
char x[] = {'h','e','\0','l','l','o','\0'}
char y[sizeof x];
memcpy(y, x, sizeof x);
Code could use "%c" to print everything, but what gets printed with '\0' varies amongst systems.
size_t i;
for (i=0; i< sizeof x; i++)
printf("%c", x[i]);
You need to print the array element wise, translating non-printable characters to whatever you want to see instead.
To print out he\0llo\0 do:
char x[] = {'h','e','\0','l','l','o','\0'};
for (size_t i = 0; i < sizeof(x); ++i)
{
if ('\0' == x[i])
{
printf("%s", "\\0");
}
else
{
printf("%c", x[i]);
}
}
Related
I don't understand Why don’t we have to print strings in for loop ? In normal cases we need to print arrays in for loop. For example, if we want to print the array of integers. It will be like this:
int a[n];
for (i = 0; i < n; i++){
printf("%d", a[i]);
}
But for strings like:
char s[100] = " Hello ";
printf("%s\n", s);
it is enough to write the name of array.
EDIT: It seems like I didnt ask my question properly as some of you wrote answers which is not related to my question.I edit my question.
Strings terminate with the empty character '\0', that's how it is possible to know when a string ends even without explicitly passing its length.
The difference is that C-style strings (which are char arrays) are zero-terminated, whereas int arrays are normally not zero terminated.
Theoretically, you could also create an int array which is zero-terminated and print that in a loop:
int a[] = {5,7,3,0};
for (i=0;a[i]!=0;i++)
{
printf("%d",a[i])
}
However, the problem with zero-terminated int arrays is that the number 0 could be a meaningful value, so you cannot be sure that it really is the end of the array when you encounter that value. With strings, however, the ASCII-Code 0 does not represent a meaningful value, so you can be reasonably sure that you have reached the end of the string.
Your example is far from analogous. %d refers to a single integer, while %s refers to an entire (but still single) string.
You are not passing the size of the array n[] to printf either - rather you are calling printf n times. You are printing one int just as you are printing one string.
The actual string length is not known a priori, rather printf iterates the string until it encounters the \0 terminator. Equivalent to:
for( int i = 0; s[i] != '\0'; i++)
{
printf( "%c", s[i] ) ;
}
because
char s[100] = " Hello ";
is equivalent to:
char s[100] = { ' ', 'H', 'e', 'l', 'l', 'o', ' ', '\0' } ;
Strings are multiple characters all at once. You cannot ask for a specific character from a string. If you want to do so, you must refer to it as an array of chars like this
for (i = 0; i < n; i++){
printf("%c", s[i]);
}
It is because you are already providing the parameter which is "%s" to the printf function which already has a definition telling it how to print a string.
Hope it helps.
One of the reason is that char only takes 1 byte of memeory and when you just press a character, the first index of array is filled up completely and it moves on to the next one till it encounters NULL character. This is not the case with integer array where the size is more than 1 byte and is machine dependent. So you cannot escape the first index by just pressing the number less than the maximum range. If you try to do this, it will store your numbers in first index only and hence a for loop is required there.
I'm new to C and I'm trying to write a program that prints the ASCII value for every letter in a name that the user enters. I attempted to store the letters in an array and try to print each ASCII value and letter of the name separately but, for some reason, it only prints the value of the first letter.
For example, if I write "Anna" it just prints 65 and not the values for the other letters in the name. I think it has something to do with my sizeof(name)/sizeof(char) part of the for loop, because when I print it separately, it only prints out 1.
I can't figure out how to fix it:
#include <stdio.h>
int main(){
int e;
char name[] = "";
printf("Enter a name : \n");
scanf("%c",&name);
for(int i = 0; i < (sizeof(name)/sizeof(char)); i++){
e = name[i];
printf("The ASCII value of the letter %c is : %d \n",name[i],e);
}
int n = (sizeof(name)/sizeof(char));
printf("%d", n);
}
Here's a corrected, annotated version:
#include <stdio.h>
#include <string.h>
int main() {
int e;
char name[100] = ""; // Allow for up to 100 characters
printf("Enter a name : \n");
// scanf("%c", &name); // %c reads a single character
scanf("%99s", name); // Use %s to read a string! %99s to limit input size!
// for (int i = 0; i < (sizeof(name) / sizeof(char)); i++) { // sizeof(name) / sizeof(char) is a fixed value!
size_t len = strlen(name); // Use this library function to get string length
for (size_t i = 0; i < len; i++) { // Saves calculating each time!
e = name[i];
printf("The ASCII value of the letter %c is : %d \n", name[i], e);
}
printf("\n Name length = %zu\n", strlen(name)); // Given length!
int n = (sizeof(name) / sizeof(char)); // As noted above, this will be ...
printf("%d", n); // ... a fixed value (100, as it stands).
return 0; // ALWAYS return an integer from main!
}
But also read the comments given in your question!
This is a rather long answer, feel free to skip to the end for the code example.
First of all, by initialising a char array with unspecified length, you are making that array have length 1 (it only contains the empty string). The key issue here is that arrays in C are fixed size, so name will not grow larger.
Second, the format specifier %c causes scanf to only ever read one byte. This means that even if you had made a larger array, you would only be reading one byte to it anyway.
The parameter you're giving to scanf is erroneous, but accidentally works - you're passing a pointer to an array when it expects a pointer to char. It works because the pointer to the array points at the first element of the array. Luckily this is an easy fix, an array of a type can be passed to a function expecting a pointer to that type - it is said to "decay" to a pointer. So you could just pass name instead.
As a result of these two actions, you now have a situation where name is of length 1, and you have read exactly one byte into it. The next issue is sizeof(name)/sizeof(char) - this will always equal 1 in your program. sizeof char is defined to always equal 1, so using it as a divisor causes no effect, and we already know sizeof name is equal to 1. This means your for loop will only ever read one byte from the array. For the exact same reason n is equal to 1. This is not erroneous per se, it's just probably not what you expected.
The solution to this can be done in a couple of ways, but I'll show one. First of all, you don't want to initialize name as you do, because it always creates an array of size 1. Instead you want to manually specify a larger size for the array, for instance 100 bytes (of which the last one will be dedicated to the terminating null byte).
char name[100];
/* You might want to zero out the array too by eg. using memset. It's not
necessary in this case, but arrays are allowed to contain anything unless
and until you replace their contents.
Parameters are target, byte to fill it with, and amount of bytes to fill */
memset(name, 0, sizeof(name));
Second, you don't necessarily want to use scanf at all if you're reading just a byte string from standard input instead of a more complex formatted string. You could eg. use fgets to read an entire line from standard input, though that also includes the newline character, which we'll have to strip.
/* The parameters are target to write to, bytes to write, and file to read from.
fgets writes a null terminator automatically after the string, so we will
read at most sizeof(name) - 1 bytes.
*/
fgets(name, sizeof(name), stdin);
Now you've read the name to memory. But the size of name the array hasn't changed, so if you used the rest of the code as is you would get a lot of messages saying The ASCII value of the letter is : 0. To get the meaningful length of the string, we'll use strlen.
NOTE: strlen is generally unsafe to use on arbitrary strings that might not be properly null-terminated as it will keep reading until it finds a zero byte, but we only get a portable bounds-checked version strnlen_s in C11. In this case we also know that the string is null-terminated, because fgets deals with that.
/* size_t is a large, unsigned integer type big enough to contain the
theoretical maximum size of an object, so size functions often return
size_t.
strlen counts the amount of bytes before the first null (0) byte */
size_t n = strlen(name);
Now that we have the length of the string, we can check if the last byte is the newline character, and remove it if so.
/* Assuming every line ends with a newline, we can simply zero out the last
byte if it's '\n' */
if (name[n - 1] == '\n') {
name[n - 1] = '\0';
/* The string is now 1 byte shorter, because we removed the newline.
We don't need to calculate strlen again, we can just do it manually. */
--n;
}
The loop looks quite similar, as it was mostly fine to begin with. Mostly, we want to avoid issues that can arise from comparing a signed int and an unsigned size_t, so we'll also make i be type size_t.
for (size_t i = 0; i < n; i++) {
int e = name[i];
printf("The ASCII value of the letter %c is : %d \n", name[i], e);
}
Putting it all together, we get
#include <stdio.h>
#include <string.h>
int main() {
char name[100];
memset(name, 0, sizeof(name));
printf("Enter a name : \n");
fgets(name, sizeof(name), stdin);
size_t n = strlen(name);
if (n > 0 && name[n - 1] == '\n') {
name[n - 1] = '\0';
--n;
}
for (size_t i = 0; i < n; i++){
int e = name[i];
printf("The ASCII value of the letter %c is : %d \n", name[i], e);
}
/* To correctly print a size_t, use %zu */
printf("%zu\n", n);
/* In C99 main implicitly returns 0 if you don't add a return value
yourself, but it's a good habit to remember to return from functions. */
return 0;
}
Which should work pretty much as expected.
Additional notes:
This code should be valid C99, but I believe it's not valid C89. If you need to write to the older standard, there are several things you need to do differently. Fortunately, your compiler should warn you about those issues if you tell it which standard you want to use. C99 is probably the default these days, but older code still exists.
It's a bit inflexible to be reading strings into fixed-size buffers like this, so in a real situation you might want to have a way of dynamically increasing the size of the buffer as necessary. This will probably require you to use C's manual memory management functionality like malloc and realloc, which aren't particularly difficult but take greater care to avoid issues like memory leaks.
It's not guaranteed the strings you're reading are in any specific encoding, and C strings aren't really ideal for handling text that isn't encoded in a single-byte encoding. There is support for "wide character strings" but probably more often you'll be handling char strings containing UTF-8 where a single codepoint might be multiple bytes, and might not even represent an individual letter as such. In a more general-purpose program, you should keep this in mind.
If we need write a code to get ASCII values of all elements in a string, then we need to use "%d" instead of "%c". By doing this %d takes the corresponding ascii value of the following character.
If we need to only print the ascii value of each character in the string. Then this code will work:
#include <stdio.h>
char str[100];
int x;
int main(){
scanf("%s",str);
for(x=0;str[x]!='\0';x++){
printf("%d\n",str[x]);
}
}
To store all corresponding ASCII value of character in a new variable, we need to declare an integer variable and assign it to character. By this way the integer variable stores ascii value of character. The code is:
#include <stdio.h>
char str[100];
int x,ascii;
int main(){
scanf("%s",str);
for(x=0;str[x]!='\0';x++){
ascii=str[x];
printf("%d\n",ascii);
}
}
I hope this answer helped you.....😊
I'm still new to programming but lets say I have a two dimensional char array with one letter in each array. Now I'm trying to combine each of these letters in the array into one array to create a word.
So grid[2][4]:
0|1|2|3
0 g|o|o|d
1 o|d|d|s
And copy grid[0][0], grid[0][1], grid[0][2], grid[0][3] into a single array destination[4] so it reads 'good'. I have something like
char destination[4];
strcpy(destination, grid[0][1]);
for(i=0; i<4; i++)
strcat(destination, grid[0][i]);
but it simply crashes..
Any step in the right direction is appreciated.
In C, the runtime library functions strcpy and strcat require zero terminated strings. What you're handing to them are not zero terminated, and so these functions will crash due to their dependency on that terminating zero to indicate when they should stop. They are running through RAM until they read a zero, which could be anywhere in RAM, including protected RAM outside your program, causing a crash. In modern work we consider functions like strcpy and strcat to be unsafe. Any kind of mistake in handing them pointers causes this problem.
Versions of strcpy and strcat exist, with slightly different names, which require an integer or size_t indicating their maximum valid size. strncat, for example, has the signature:
char * strncat( char *destination, const char *source, size_t num );
If, in your case, you had used strncat, providing 4 for the last parameter, it would not have crashed.
However, an alternative exists you may prefer to explore. You can simply use indexing, as in:
char destination[5]; // I like room for a zero terminator here
for(i=0; i<4; i++)
destination[i] = grid[0][i];
This does not handle the zero terminator, which you might append with:
destination[4] = 0;
Now, let's assume you wanted to continue, putting both words into a single output string. You might do:
char destination[10]; // I like room for a zero terminator here
int d=0;
for(r=0; r<2; ++r ) // I prefer the habit of prefix instead of postfix
{
for( i=0; i<4; ++i )
destination[d++] = grid[r][i];
destination[d++] = ' ';// append a space between words
}
Following whatever processing is required on what might be an ever larger declaration for destination, append a zero terminator with
destination[ d ] = 0;
strcpy copies strings, not chars. A string in C is a series of chars, followed by a \0. These are called "null-terminated" strings. So your calls to strcpy and strcat aren't giving them the right kind of parameters.
strcpy copies character after character until it hits a \0; it doesn't just copy the one char you're giving it a pointer to.
If you want to copy a character, can just assign it.
char destination[5];
for(i = 0; i < 4; i++)
destination[i] = grid[0][i];
destination[i] = '\0';
I was just printing some characters in C.
I have declared char c and assigned its characters.
then using a for loop i try to print the characters one by one.
I have used pointers, of course.
#include <stdio.h>
#include <string.h>
int main()
{
char c[4] = {"hia"};
int i;
for(i=0;i<4;i++)
{
printf(&c[i]);
}
return 0;
}
However when I compile my code using turbo, i get output "hiaiaa" instead of "hia"! What am i doing wrong here?
Your printf() call is broken. You are using the string (from the point you specify) as the formatting string. This will not print single characters. Instead each call will print from where its formatting string starts, to end of the string.
This means the first call will print all of c, the next will print from c[1] and onwards, and so on. Not at all what you wanted.
If you want to print single characters, use the %c format specifier:
printf("%c", c[i]);
No pointer necessary, since the character is passed by value to printf().
This is what happened in your loop:
0. hia
1. ia
2. a
3. \0
However, you want to print exactly one char at a time, not a null terminated string, so you should pass it as char not a char*:
printf( "%c", c[i] )
Also, you are looping four times, but string length is just three. You should use:
for( i = 0; i < strlen( c ); i++ )
...
The printf function have an char* as first argument, that's correct. However, it prints a string (that is, a zero-terminated sequence of char) so it will always do that.
If you want to print one character at a time, then you have to use that format, like in:
printf("%c\n", c[i]);
You also have another problem, and that is that you try to print the zero terminator as well. This character is not printable so will not show. Use e.g. i < strlen(c) as the loop condition to overcome this.
Also, instead of printing character-by-character, print it all as one string:
printf("%s\n", c);
1) For loop size should i<3 , not i<4 (i=3 refers to the null character at the end of the string)
2) use printf("%c",c[i]);
Explanation of what you're seeing: In each loop, printf is printing a null-terminated string. This string starts in every loop one char later inside your array.
How it should be done, depends on what you're intending. If you want to print the string char by char via pointer you may use:
char *p=&c[0];
while (*p) {
printf("%c", *p);
p++;
}
Your question is to print string using pointer. You could use
printf("%s", c);
or character by character as (include library string.h for this)
for(i=0;i<strlen(c);i++)
{
printf("%c", c[i]);
}
in C strings are stored as character arrays and are terminated by a zero-value, so called zero-terminated strings. Btw, this is why you have to make the array size of 4 for thee real chars.
In your example, you are passing pointers th each char to the printf function and printf prints the strings from your pointer to the next null-value . The 1st pass prints "hia", the 2nd ia and the 3rd a.
To print a single char in each pass, you have to use
printf ("%c", c[i]);
Your loop will call printf with the following parameter:
printf("hia"); // first loop iteration
printf("ia"); // second loop iteration
printf("a"); // third loop iteration
printf(""); // fourth loop iteration
You probably meant to print one character at a time:
for(i=0;i<3;i++) // No need to print the string termination character.
{
printf("%c", c[i]); // "%c" is the printf format code to print a single character
}
In my C program I am trying to copy an array of char's to another array whilst removing the first element (element 0).
I have written:
char array1[9];
char array2[8];
int i, j;
for(i = 1, j = 0 ; i < 10, j < 9; i++, j++){
array2[j] = array1[i];
}
printf(array2);
When I print array2, it gives me a stack overflow.
Any ideas?
Two issues:
First, when printing a string with printf, and working with other standard C string functions in general, your char arrays need to be null-terminated so the functions know where the string ends. You are also writing one past the end of your arrays.
Second, when using printf, it is almost always a bad idea to use the string you want to print as the format string. Use
printf("%s", array2);
instead. If you use printf as in the original example and array2 can be influenced by the user, then your program is likely vulnerable to a format string vulnerability.
Use memcpy():
memcpy( array2, &array1[1], 8 );
Thats easier.
Your string isn't null-terminated, so when its printed it continues printing characters past the 8 you've allocated looking for one but runs out of stack space before then. You're also writing to one character more than you've allocated and your conditions should be "combined" with && -- a , ignores the result of the first expression. You should also avoid using a string variable as the string formatter to printf.
Here's your code fixed:
char array1[10] = "123456789";
char array2[9];
int i, j;
for(i = 1, j = 0 ; i < 10 && j < 9; i++, j++){
array2[j] = array1[i];
}
printf("%s\n", array2);
You can also simplify the loop by using a single index variable i and indexing array2 with i+. You can also remove the loop entirely by using strncpy, but be aware that if n is less than the length of the string + 1 it won't add a null-terminator.
It's not necessary to use an extra array2 like
printf("%.8s",array1+1);
When you say printf(array2), it thinks it's printing a null-terminated string. Since there is (possibly) no \0 in array2, printf continues on past the end of array2, wandering into memory it isn't supposed to.
To further expand on marcog's answer: you are declaring array1 with 9 elements, 0-8, and then writing from 0-9 (10 elements). Same thing with array2.
Just use strcpy() (if they're both strings!) strcpy() wants a pointer to the source and a pointer to the destination. If you want to skip the first element of the source array just pass source + 1:
char source[] = "ffoo";
char dest[] = "barbar";
strcpy(dest, source + 1);
// now dest is "foo" (since the ending \0 is copied too)
printf("\n%s\n", dest);