How to iterate over a string in C?

问题:

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?

回答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++){

回答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.

回答3:

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.

回答4:

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.

回答5:

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]);

    }


 }

回答6:

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;
}

回答7:

• 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.

回答8:

Replace sizeof with strlen and it should work.

回答9:

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.

回答10:

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]);

}

回答11:

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).