在不使用 string.h 库的情况下计算 C 中字符串的长度（使用 char 指针）

Question

I need to count the length of a string in C without using the string.h library, using char pointers. So far all the methods I've seen use the string.h library. Is it possible to do it without i?

Answer 1

Yes it is possible, the end of a string in C is usually null terminated, so you can loop through the string until you find the '\0' character, counting each as you go.

Answer 2

When talking about strings in C, you're essentially talking about an array of char , terminated by the nil character ( '\0' ). The length of a string, therefore is:

size_t len(const char *s) {
    // if no string was passed (null pointer), just return
    if (s == NULL)
        return 0;

    size_t ln;
    // iterate over string until you encounter the terminating char
    for (ln=0;s[ln] != '\0'; ln++);

    return ln;
}

As pointed out by @chux-ReinstateMonica: the correct type to return here is size_t , if you want to avoid potential overflow issues + you're looking for a drop-in replacement for the strlen() function in strings.h . The length of a string can, after all, never be negative, so using the unsigned size_t makes more sense than the signed int .

Now this obviously assumes the input was a valid string. If I were to do something like:

int main( void ) {
    char * random = malloc(1024); // allocate 1k chars, without initialising memory
    printf("length is: %zu\n", len(random)); // use %zu because size_t is unsigned!
    return 0;
}

There's no guarantee the memory I allocated will even contain a terminating character, and therefore it's not impossible for the function to return a value greater than 1024. Same goes for something like:

char foo[3] = {'f', 'o', 'o'}; // full array, no terminating char

This string, in memory, could look something like this:

//the string - random memory after the array
| f | o | o | g | a | r | b | a | g | e | U | B | \0 |

Making it look like a string of 12 characters. If you then use the assumed length of 12 to write past the actual array, all bets are off (undefined behaviour).

This string should've been:

char foo[4] = {'f', 'o', 'o', '\0'}; // or char[4] foo = "foo";

You can read an old answer I gave elsewhere for some more details on how to get the length of a string passed to a function here

Answer 3

Here example:

size_t pos(char* s)
{
    size_t lenS; //len of string

    for (lenS = 0; s[lenS]; lenS++);

    return lenS;
}