How to get the string size in bytes? - c

As the title implies, my question is how to get the size of a string in C. Is it good to use sizeof if I've declared it (the string) in a function without malloc in it? Or, if I've declared it as a pointer? What if I initialized it with malloc? I would like to have an exhaustive response.

You can use strlen. Size is determined by the terminating null-character, so passed string should be valid.
If you want to get size of memory buffer, that contains your string, and you have pointer to it:
If it is dynamic array(created with malloc), it is impossible to get
it size, since compiler doesn't know what pointer is pointing at.
(check this)
If it is static array, you can use sizeof to get its size.
If you are confused about difference between dynamic and static arrays, check this.

Use strlen to get the length of a null-terminated string.
sizeof returns the length of the array not the string. If it's a pointer (char *s), not an array (char s[]), it won't work, since it will return the size of the pointer (usually 4 bytes on 32-bit systems). I believe an array will be passed or returned as a pointer, so you'd lose the ability to use sizeof to check the size of the array.
So, only if the string spans the entire array (e.g. char s[] = "stuff"), would using sizeof for a statically defined array return what you want (and be faster as it wouldn't need to loop through to find the null-terminator) (if the last character is a null-terminator, you will need to subtract 1). If it doesn't span the entire array, it won't return what you want.
An alternative to all this is actually storing the size of the string.

While sizeof works for this specific type of string:
char str[] = "content";
int charcount = sizeof str - 1; // -1 to exclude terminating '\0'
It does not work if str is pointer (sizeof returns size of pointer, usually 4 or 8) or array with specified length (sizeof will return the byte count matching specified length, which for char type are same).
Just use strlen().

If you use sizeof()then a char *str and char str[] will return different answers. char str[] will return the length of the string(including the string terminator) while char *str will return the size of the pointer(differs as per compiler).

I like to use:
(strlen(string) + 1 ) * sizeof(char)
This will give you the buffer size in bytes. You can use this with snprintf() may help:
const char* message = "%s, World!";
char* string = (char*)malloc((strlen(message)+1))*sizeof(char));
snprintf(string, (strlen(message)+1))*sizeof(char), message, "Hello");
Cheers! Function: size_t strlen (const char *s)

There are two ways of finding the string size bytes:
1st Solution:
# include <iostream>
# include <cctype>
# include <cstring>
using namespace std;
int main()
{
char str[] = {"A lonely day."};
cout<<"The string bytes for str[] is: "<<strlen(str);
return 0;
}
2nd Solution:
# include <iostream>
# include <cstring>
using namespace std;
int main()
{
char str[] = {"A lonely day."};
cout<<"The string bytes for str[] is: "<<sizeof(str);
return 0;
}
Both solution produces different outputs. I will explain it to you after you read these.
The 1st solution uses strlen and based on cplusplus.com,
The length of a C string is determined by the terminating null-character: A C string is as long as the number of characters between the beginning of the string and the terminating null character (without including the terminating null character itself).
That can explain why does the 1st Solution prints out the correct string size bytes when the 2nd Solution prints the wrong string size bytes. But if you still don't understand, then continue reading.
The 2nd Solution uses sizeof to find out the string size bytes. Based on this SO answer, it says (modified it):
sizeof("f") must return 2 string size bytes, one for the 'f' and one for the terminating '\0' (terminating null-character).
That is why the output is string size bytes 14. One for the whole string and one for '\0'.
Conclusion:
To get the correct answer for 2nd Solution, you must do sizeof(str)-1.
References:
Sizeof string literal
https://cplusplus.com/reference/cstring/strlen/?kw=strlen

Related

Question about strcpy using char pointer in C

I'm doing some programming with C, and I have a minor problem using strcpy.
char* file="It has something inside"
int size= sizeof(file);
char* file_save = malloc(sizeof(char)*size);
strcpy(file_save,file);
My code stopped working in the last line.
What can be the problem here?
It seems like something went wrong outside this part of the code. It works perfectly well if I change sizeof into strlen in the online gdb, but it still stops in the strcpy line on my computer's VS code. Thank you for helping me.
As comments suggest, the size is the size of variable char pointer which is probably 8 if you are working on a 64-bit machine. So you need the size of string the file is pointing to. You can get that like below:
int size = strlen(file) + 1;
The strlen returns the number of bytes this string has. Every string is finished by a null terminator \0. While you are writing in a file it doesn't matter which bytes are for the string and which is the null terminator so you need to count both.
The variable file has the pointer type char *
char* file="It has something inside";
It would be even better to declare the pointer with the qualifier const because you may not change the string literal pointed to by the pointer
const char *file = "It has something inside";
Its size returned by the operator sizeof
int size= sizeof(file);
does not depend on what string the pointer points to and usually is equal to 4 or 8 bytes.
As you are going to copy the pointed string literal you need to determine its length. To do this you can use standard string function strlen. It returns the number of characters in a string until the terminating zero character '\0' is encountered.
So you need to write
size_t size = strlen( file );
Now you need to allocate memory for size characters plus one for the terminating zero character '\0' of the string literal.
char* file_save = malloc( size + 1 );
So all is ready to copy the source string provided that the memory was allocated successfully.
strcpy( file_save, file );
What can be the problem here?
A pointer is not an array and not a string. An array is not a pointer. A string is not a pointer.
Below, file is a pointer.
char* file = "It has something inside";
int size = sizeof(file);
sizeof(file) is the size of a pointer, not the size of array nor the size of a string.
Later code needs to use the size of a string, not the size of a pointer.
To find the size of a string, use the pointer file, which points to a string and then call strlen().
// `length` is the string length: characters up to, but not including the null character.
size_t length = strlen(file);
// `size` is the string size: characters up to, and including the null character.
size_t size = length + 1;
Use size to determine allocation needs for a copy of the string.

Getting wrong string length

I am trying to get the length of a string but i am getting the wrong value, it is saying that it is only 4 characters long. Why is this? am i using sizeof() correctly?
#include <stdio.h>
int main(void)
{
char *s;
int len;
s = "hello world";
len = sizeof(s);
printf("%d\n", len);
}
The sizeof operator is returning the size of the pointer. If you want the length of a string, use the strlen function.
Even if you had an array (e.g. char s[] = "hello world") the sizeof operator would return the wrong value, as it would return the length of the array which includes the string terminator character.
Oh and as a side note, if you want a string pointer to point to literal string, you should declare it const char *, as string literals are constant and can't be modified.
You have declared s as a pointer. When applied to a pointer, sizeof() returns the size of the pointer, not the size of the element pointed to. On your system, the size of a pointer to char happens to be four bytes. So you will see 4 as your output.
In addition to strlen(), you can assign string literal to array of chars
char s[] = "hello world", in this case sizeof() returns size of array in bytes. In this particular case 12, one extra byte for \0 character at the end of the string.
Runtime complexity of sizeof() is O(1).
Complexity of strlen() is O(n).

C Noob: Define size of a char array while copying contents of char* into it

I have a basic question in C.
I need to print the contents of a char pointer. The contents are binary and therefore I use hex format to see the contents.
Would detecting a null still work?
unsigned char *input = "������";
printf("input =");
int count = 0;
while(*input != '\0'){
printf("%02x", *input);
input++;
}
printf("\n");
Now what happens if I have to copy the pointer to a char array?
How can I assign the size of the char array? I understand sizeof returns only the size of datatype that char points to. But is there any way?
unsigned char copyInput[size??];
strcpy(copyInput, input);
for (i=0, i <size?, i++)
{
printf("copyInput[%d]= %02x", i, copyInput[i]);
}
Thanks in advance!
1) To the extent that C has strings at all, they are defined as "an arbitrary contiguous sequence of nonzero bytes, terminated with a zero byte". Therefore, if your binary data is guaranteed never to contain bytes whose value is zero, you can safely treat it as a C string (use the str* functions with it, etc). But if your binary data might have zero bytes somewhere in the middle, you need to track the length separately and operate on it with the mem* functions instead.
2) You use strlen to find the length of the string (without the terminating zero byte). However, in standard C89 you can't use the result of strlen to set the size of a char[] variable, because the size has to be known at compile time. If you're using C99 or GNU extensions, you can define the size of an array at runtime:
size_t n = strlen(s1);
char s2[n+1];
memcpy(s2, s1, n);
The n+1 is necessary, or you won't have space for the terminating NUL. If you can't use C99 nor GNU extensions, your only option is to allocate space on the heap:
size_t n = strlen(s1);
char *s2 = malloc(n+1);
memcpy(s2, s1, n);
or, with a common library extension, just
char *s2 = strdup(s1);
Either way, don't forget to free(s2) later. By the way, this is a case where it would have been safe to use strcpy, because you know by construction that the destination buffer is big enough. I used memcpy because it may be slightly more efficient and it means human readers won't see "strcpy" and start worrying.
If it's a bunch of chars terminated with a 0, just use strlen() since that is C's definition of a string. It doesn't matter than some (or most) of the characters might be unprintable, as long as 0 is the terminator.
You will have problems if any of the input bytes are 0. In this case the loop will stop at that character. Otherwise, you can treat it as a string.
Treating it as a string, you can use strlen() to get the input's size and then dynamically allocate memory to your copy. The copy can be made with strcpy as you did, but it is safer to use strncpy.
char *input = "input binary array";
int count = strlen(input)+1; // plus '\0'
char *copy = (char *) malloc(count*sizeof(char));
strncpy(copy, input, count+1);

Trying to Wrap My Head Around String Sizes in C

A friend and I are doing a C programming unit for college.
We understand that there is no "string" per se in C, and instead, a string is defined by being an array of characters. Awesome!
So when dealing with "strings" is obvious that a proper understanding arrays and pointers is important.
We were doing really well understanding pointer declaration, when and when not to dereference the pointer, and played around with a number of printf's to test our experiments. All with great success.
However, when we used this:
char *myvar = "";
myvar = "dhjfejfdhdkjfhdjkfhdjkfhdjfhdfhdjhdsjfkdhjdfhddskjdkljdklc";
printf("Size is %d\n", sizeof(myvar));
and it spits out Size is 8!
Why 8? Clearly there are more than 8 bytes being consumed by 'myvar' (or is it)?
(I should be clear and point out that I am VERY aware of 'strlen'. This is not an exercise in getting the length of a string. This is about trying to understand why sizeof returns 8 bytes for the variable myvar.)
8 is the size of the pointer.
myvar is a pointer to char (hence char*) and in 64 bit system pointers are 64 bit = 8 byte
To get size of a null-terminated string use this code :
#include<string.h>
#include<stdio.h>
int main()
{
char *x="hello there";
printf("%d\n",strlen(x));
return 0;
}
Well as AbiusX said, the reason why sizeof is returning 8 is because you are finding the size of a pointer (and I'm guessing you're on a 64-bit machine). For example, that same code-snippet would return 4 on my machine.
Strings in C are kept as an array of characters followed by a null terminator. So when you do this...
const char *message = "hello, world!"
It's actually stored in memory as:
'h''e''l''l''o'','' ''w''o''r''l''d''!''\0'...garbage here
If you read past the null terminator, you'll likely just find whatever garbage happens to be in memory there at the time. So in order to find the length of a string in C, you need to start at the beginning of the string and read until the null terminator.
size_t count = 0;
const char *message = "hello, world!";
for ( ; message[count] != '\0'; count++ );
printf("size of message %u\n", count);
Now this is an O(n) operation (because you have to iterate over the entire array to get the size). Most higher level languages have their upper level abstraction of strings as something similar to...
struct string {
char *c_str;
size_t length;
};
And then they just keep track of how long the string is whenever they do an operation on it. This greatly speeds up finding the length of a string, which is a very common operation.
Now there is one way you can figure out the length of a string using sizeof, but I don't suggest it. Using sizeof on an array (not a pointer!) will return the size of the array multiplied by the data type size. And C can auto-figure out the size of an array as long as it can be figured out at compile-time.
const char message[] = "hello, world!";
printf("size of message %u\n", sizeof(message));
That will print the correct size of the message. Remember, this is NOT suggested. Notice that this will print one greater than the number of characters in the string. That's because it also counts the null terminator (as it has to allocate an array large enough to have the null terminator). So it's not really the real length of the string (you can always just subtract one).
myvar is a pointer. You seem to be on a 64-bit machine, so sizeof returns 8 byte in size. What you're probably looking for instead is strlen().
Like AbiusX said, 8 is the size of the pointer. strlen can tell you the length of the string (man page).

I'm new to C, can someone explain why the size of this string can change?

I have never really done much C but am starting to play around with it. I am writing little snippets like the one below to try to understand the usage and behaviour of key constructs/functions in C. The one below I wrote trying to understand the difference between char* string and char string[] and how then lengths of strings work. Furthermore I wanted to see if sprintf could be used to concatenate two strings and set it into a third string.
What I discovered was that the third string I was using to store the concatenation of the other two had to be set with char string[] syntax or the binary would die with SIGSEGV (Address boundary error). Setting it using the array syntax required a size so I initially started by setting it to the combined size of the other two strings. This seemed to let me perform the concatenation well enough.
Out of curiosity, though, I tried expanding the "concatenated" string to be longer than the size I had allocated. Much to my surprise, it still worked and the string size increased and could be printf'd fine.
My question is: Why does this happen, is it invalid or have risks/drawbacks? Furthermore, why is char str3[length3] valid but char str3[7] causes "SIGABRT (Abort)" when sprintf line tries to execute?
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void main() {
char* str1 = "Sup";
char* str2 = "Dood";
int length1 = strlen(str1);
int length2 = strlen(str2);
int length3 = length1 + length2;
char str3[length3];
//char str3[7];
printf("%s (length %d)\n", str1, length1); // Sup (length 3)
printf("%s (length %d)\n", str2, length2); // Dood (length 4)
printf("total length: %d\n", length3); // total length: 7
printf("str3 length: %d\n", (int)strlen(str3)); // str3 length: 6
sprintf(str3, "%s<-------------------->%s", str1, str2);
printf("%s\n", str3); // Sup<-------------------->Dood
printf("str3 length after sprintf: %d\n", // str3 length after sprintf: 29
(int)strlen(str3));
}
This line is wrong:
char str3[length3];
You're not taking the terminating zero into account. It should be:
char str3[length3+1];
You're also trying to get the length of str3, while it hasn't been set yet.
In addition, this line:
sprintf(str3, "%s<-------------------->%s", str1, str2);
will overflow the buffer you allocated for str3. Make sure you allocate enough space to hold the complete string, including the terminating zero.
void main() {
char* str1 = "Sup"; // a pointer to the statically allocated sequence of characters {'S', 'u', 'p', '\0' }
char* str2 = "Dood"; // a pointer to the statically allocated sequence of characters {'D', 'o', 'o', 'd', '\0' }
int length1 = strlen(str1); // the length of str1 without the terminating \0 == 3
int length2 = strlen(str2); // the length of str2 without the terminating \0 == 4
int length3 = length1 + length2;
char str3[length3]; // declare an array of7 characters, uninitialized
So far so good. Now:
printf("str3 length: %d\n", (int)strlen(str3)); // What is the length of str3? str3 is uninitialized!
C is a primitive language. It doesn't have strings. What it does have is arrays and pointers. A string is a convention, not a datatype. By convention, people agree that "an array of chars is a string, and the string ends at the first null character". All the C string functions follow this convention, but it is a convention. It is simply assumed that you follow it, or the string functions will break.
So str3 is not a 7-character string. It is an array of 7 characters. If you pass it to a function which expects a string, then that function will look for a '\0' to find the end of the string. str3 was never initialized, so it contains random garbage. In your case, apparently, there was a '\0' after the 6th character so strlen returns 6, but that's not guaranteed. If it hadn't been there, then it would have read past the end of the array.
sprintf(str3, "%s<-------------------->%s", str1, str2);
And here it goes wrong again. You are trying to copy the string "Sup<-------------------->Dood\0" into an array of 7 characters. That won't fit. Of course the C function doesn't know this, it just copies past the end of the array. Undefined behavior, and will probably crash.
printf("%s\n", str3); // Sup<-------------------->Dood
And here you try to print the string stored at str3. printf is a string function. It doesn't care (or know) about the size of your array. It is given a string, and, like all other string functions, determines the length of the string by looking for a '\0'.
Instead of trying to learn C by trial and error, I suggest that you go to your local bookshop and buy an "introduction to C programming" book. You'll end up knowing the language a lot better that way.
There is nothing more dangerous than a programmer who half understands C!
What you have to understand is that C doesn't actually have strings, it has character arrays. Moreover, the character arrays don't have associated length information -- instead, string length is determined by iterating over the characters until a null byte is encountered. This implies, that every char array should be at least strlen + 1 characters in length.
C doesn't perform array bounds checking. This means that the functions you call blindly trust you to have allocated enough space for your strings. When that isn't the case, you may end up writing beyond the bounds of the memory you allocated for your string. For a stack allocated char array, you'll overwrite the values of local variables. For heap-allocated char arrays, you may write beyond the memory area of your application. In either case, the best case is you'll error out immediately, and the worst case is that things appear to be working, but actually aren't.
As for the assignment, you can't write something like this:
char *str;
sprintf(str, ...);
and expect it to work -- str is an uninitialized pointer, so the value is "not defined", which in practice means "garbage". Pointers are memory addresses, so an attempt to write to an uninitialized pointer is an attempt to write to a random memory location. Not a good idea. Instead, what you want to do is something like:
char *str = malloc(sizeof(char) * (string length + 1));
which allocates n+1 characters worth of storage and stores the pointer to that storage in str. Of course, to be safe, you should check whether or not malloc returns null. And when you're done, you need to call free(str).
The reason your code works with the array syntax is because the array, being a local variable, is automatically allocated, so there's actually a free slice of memory there. That's (usually) not the case with an uninitialized pointer.
As for the question of how the size of a string can change, once you understand the bit about null bytes, it becomes obvious: all you need to do to change the size of a string is futz with the null byte. For example:
char str[] = "Foo bar";
str[1] = (char)0; // I'd use the character literal, but this editor won't let me
At this point, the length of the string as reported by strlen will be exactly 1. Or:
char str[] = "Foo bar";
str[7] = '!';
after which strlen will probably crash, because it will keep trying to read more bytes from beyond the array boundary. It might encounter a null byte and then stop (and of course, return the wrong string length), or it might crash.
I've written all of one C program, so expect this answer to be inaccurate and incomplete in a number of ways, which will undoubtedly be pointed out in the comments. ;-)
Your str3 is too short - you need to add extra byte for null-terminator and the length of "<-------------------->" string literal.
Out of curiosity, though, I tried
expanding the "concatenated" string to
be longer than the size I had
allocated. Much to my surprise, it
still worked and the string size
increased and could be printf'd fine.
The behaviour is undefined so it may or may not segfault.
strlen returns the length of the string without the trailing NULL byte (\0, 0x00) but when you create a variable to hold the combined strings you need to add that 1 character.
char str3[length3 + 1];
…and you should be all set.
C strings are '\0' terminated and require an extra byte for that, so at least you should do
char str3[length3 + 1]
will do the job.
In sprintf() ypu are writing beyond the space allocated for str3. This may cause any type of undefined behavior (If you are lucky then it will crash). In strlen(), it is just searching for a NULL character from the memory location you specified and it is finding one in 29th location. It can as well be 129 also i.e. it will behave very erratically.
A few important points:
Just because it works doesn't mean it's safe. Going past the end of a buffer is always unsafe, and even if it works on your computer, it may fail under a different OS, different compiler, or even a second run.
I suggest you think of a char array as a container and a string as an object that is stored inside the container. In this case, the container must be 1 character longer than the object it holds, since a "null character" is required to indicate the end of the object. The container is a fixed size, and the object can change size (by moving the null character).
The first null character in the array indicates the end of the string. The remainder of the array is unused.
You can store different things in a char array (such as a sequence of numbers). It just depends on how you use it. But string function such as printf() or strcat() assume that there is a null-terminated string to be found there.

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