查找字符串C中字符的第一个位置

Question

Find the first position of a character c in string

Here is my code of the function

int char_index(int c, char *string) {
    int flag = 0;
    int i = 0;
    int index = 0;

    for(i = 0; string[i] != '\0'; i++){
        if (string[i] == c){
            flag++;
        }
        if (flag == 0){
            index = NOT_IN_STRING;
        } 
        else {
            index = i+1;
        }
    }
    return index;
}

The function is expected to return the position of the character, if the character doesn't exist, the return should be : NOT_IN_STRING

Answer 1

You do not interrupt the loop even when the target character is found in the string. So the length of the string will be returned as an index of the target character.

Also it is better when the function returns an object of type size_t instead of int .

And the function declaration should look like

size_t char_index( const char *s, char c );

that is the pointer shall have the qualifier const because the string is not changed inside the function.

Take into account that the C standard includes almost a similar function strchr that is declared like

char * strchr( const char *s, char c );

Here is a demonstrative program that shows how the function can be implemented.

#include <stdio.h>

#define NOT_IN_STRING   ( size_t )-1

size_t char_index( const char *s, char c )
{
    size_t i = 0;

    while ( s[i] != '\0' && s[i] != c ) ++i;

    return s[i] != '\0' ? i : NOT_IN_STRING;
}

int main( void )
{
    const char *s = "Betty";

    for ( size_t i = 0; s[i] != '\0'; i++ )
    {
        printf( "%c: %zu\n", s[i], char_index( s, s[i] ) );
    }
}

The program output is

B: 0
e: 1
t: 2
t: 2
y: 4

The shown function excludes the terminating zero from searching. If you want to include the terminating zero in searching then just change the return statement of the function the folloing way

return s[i] == c ? i : NOT_IN_STRING;

Answer 2

Consider using char *strchr(const char *string, int c); from standard library.

Anyway the function should be done in this way:

int char_index(char c, char *string) {
    for (int i = 0; string[i] != '\0'; i++)
        if (string[i] == c)
            return i;

    return NOT_IN_STRING;
}

Answer 3

Don't use all those indexes. It's too difficult to read.

/* Find first occurence of a character in a string. */
#include <stdio.h>

#define NOT_IN_STRING -1

int
char_index(int c, const char *string)
{
        const char *start = string;
        while( *string != c && *string ) {
                string++;
        }
        return ( *string == c) ? string - start : NOT_IN_STRING;
}

int
main(int argc, char **argv)
{
        const char *needle = argc > 1 ? argv[1] : "";
        const char *haystack = argc > 2 ? argv[2] : "abcdefgh";

        printf("Target string: %s\n", haystack);
        while(*needle) {
                printf("\t%c: %d\n", *needle, char_index((int)*needle, haystack));
                needle += 1;
        }
        return 0;
}

Answer 4

Here you have another option:

int instring(int c, const char *str)
{
    int result = NOT_IN_STRING;
    const char savedstr = str;

    while(*str)
    {
        if(*str == c)
        {
            result = str - savedstr;
            break;
        }
        str++;
    }
    return result;
}

and here the comparison https://godbolt.org/z/uCBRm2

Answer 5

Couple of problems in this code. First the flag, if character is found in the string so flag++ but what happens when the character shows up in the string more than once? the flag will be 2,3 etc.. that way you will have only the last index in string that character was. To solve that add to for loop str[i] && !flag . In addition you are checking if a char is equal to int without any casting ( str[i] == c ) change it by function atoi() or regular, simple, casting. Good Luck! James