How to iterate over a string in C?

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.

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;
}

An optimized approach:

for (char character = *string; character != '\0'; character = *++string)
{
    putchar(character); // Do something with character.
}

Most C strings are null-terminated, meaning that as soon as the character becomes a '\0' the loop should stop. The *++string is moving the pointer one byte, then dereferencing it, and the loop repeats.

The reason why this is more efficient than strlen() is because strlen already loops through the string to find the length, so you would effectively be looping twice (one more time than needed) with strlen().

sizeof(source) returns the number of bytes required by the pointer char*. You should replace it with strlen(source) which will be the length of the string you're trying to display.

Also, you should probably replace printf("%s",source[i]) with printf("%c",source[i]) since you're displaying a character.

This should work

 #include <stdio.h>
 #include <string.h>

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

    char *source = "This is an example.";
    int length = (int)strlen(source); //sizeof(source)=sizeof(char *) = 4 on a 32 bit implementation
    for (int i = 0; i < length; i++) 
    {

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

    }


 }

1. sizeof() includes the terminating null character. You should use strlen() (but put the call outside the loop and save it in a variable), but that's probably not what's causing the exception.
2. you should use "%c", not "%s" in printf - you are printing a character, not a string.

Just change sizeof with strlen.

Like this:

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

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

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

}

• 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.

The last index of a C-String is always the integer value 0, hence the phrase "null terminated string". Since integer 0 is the same as the Boolean value false in C, you can use that to make a simple while clause for your for loop. When it hits the last index, it will find a zero and equate that to false, ending the for loop.

for(int i = 0; string[i]; i++) { printf("Char at position %d is %c\n", i, string[i]); }

Replace sizeof with strlen and it should work.

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).

sizeof(source) returns sizeof a pointer as source is declared as char *. Correct way to use it is strlen(source).

Next:

printf("%s",source[i]); 

expects string. i.e %s expects string but you are iterating in a loop to print each character. Hence use %c.

However your way of accessing(iterating) a string using the index i is correct and hence there are no other issues in it.