How to iterate over a string in C?

Question

Right now I'm trying this:

#include <stdio.h>

int main(int argc, char *argv[]) {

    if (argc != 3) {

        printf("Usage: %s %s sourcecode input", argv[0], argv[1]);
    }
    else {
        char source[] = "This is an example.";
        int i;

        for (i = 0; i < sizeof(source); i++) {

            printf("%c", source[i]);
        }
    }

    getchar();

    return 0;
}

This does also NOT work:

char *source = "This is an example.";
int i;

for (i = 0; i < strlen(source); i++){

    printf("%c", source[i]);
}

I get the error

Unhandled exception at 0x5bf714cf (msvcr100d.dll) in Test.exe: 0xC0000005: Access violation while reading at position 0x00000054.

(loosely translated from german)

So what's wrong with my code?

Answer 1

You want:

for (i = 0; i < strlen(source); i++){

sizeof gives you the size of the pointer, not the string. However, it would have worked if you had declared the pointer as an array:

char source[] = "This is an example.";

but if you pass the array to function, that too will decay to a pointer. For strings it's best to always use strlen. And note what others have said about changing printf to use %c. And also, taking mmyers comments on efficiency into account, it would be better to move the call to strlen out of the loop:

int len = strlen( source );
for (i = 0; i < len; i++){

or rewrite the loop:

for (i = 0; source[i] != 0; i++){

Answer 2

• sizeof(source) is returning to you the size of a char*, not the length of the string. You should be using strlen(source), and you should move that out of the loop, or else you'll be recalculating the size of the string every loop.
• By printing with the %s format modifier, printf is looking for a char*, but you're actually passing a char. You should use the %c modifier.

Answer 3

Rather than use strlen as suggested above, you can just check for the NULL character:

#include <stdio.h>

int main(int argc, char *argv[])
{
    const char *const pszSource = "This is an example.";
    const char *pszChar = pszSource;

    while (pszChar != NULL && *pszChar != '\0')
    {
        printf("%s", *pszChar);
        ++pszChar;
    }

    getchar();

    return 0;
}

Answer 4

Replace sizeof with strlen and it should work.

Answer 5

You need a pointer to the first char to have an ANSI string.

printf("%s", source + i);

will do the job

Plus, of course you should have meant strlen(source), not sizeof(source).