How to iterate over a string in C?

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?

Solution 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++) {

Solution 2:

One common idiom is:

char* c = source;
while (*c) putchar(*c++);

A few notes:

• In C, strings are null-terminated. You iterate while the read character is not the null character.
• *c++ increments c and returns the dereferenced old value of c.
• printf("%s") prints a null-terminated string, not a char. This is the cause of your access violation.