C-将子字符串转换为int

Question

At the moment I got this simple code in C to convert days, minutes and seconds to seconds only:

EDITED (I understood the problem with atoi, like this is corrected right?):

#include <stdio.h>
#include <stdlib.h>

int getseconds(char * time)
{
    int seconds=0, i=0;
    char buffer[3];
    while (*time != '\0')
    {
        switch (*time)
        {
            case 'h': buffer[i]='\0';i=0;seconds=seconds+atoi(buffer)*3600;break;
            case 'm': buffer[i]='\0';i=0;seconds=seconds+atoi(buffer)*60;break;
            case 's': buffer[i]='\0';i=0;seconds=seconds+atoi(buffer);break;
            case ' ':break;
            default: buffer[i]=*time;i++;break;
        }
        time++;
    }
    return seconds;
}

int main()
{

    char *time = "12h  4m 58s";
    int seconds = getseconds(time);
    printf("%d",seconds);
    return 0;
}

This is working as I want, but there isn't another way to do it, without creating more variables (like C# where I just need to convert "inline". Does C only have functions that convert to variables and not to "inline")?

C# Example:

        string time = "12h 34m 58s";
        int seconds = int.Parse(time.Substring(0, 2)) * 3600 + int.Parse(time.Substring(4, 2)) * 60 + int.Parse(time.Substring(8, 2));

You can spot the number of lines difference I guess :).

Answer 1

You can use atoi() or strtol() to do this job. You do not need to extract substrings first, because these functions stop at the first character that cannot be converted. You just need to take a varying view of where the string starts. Thus, this code is fairly analogous to your C# example:

int getseconds(const char * time) {
    return atoi(time) * 3600 + atoi(time + 4) * 60 + atoi(time + 8);
}

Of course it will break on some malformed inputs, but so will the C# (albeit not all the same inputs).

Answer 2

sscanf() with "%n" to detect the end of scanning is close but does use more variables.

int getseconds2(const char * time) {
   int h,m,s;
   int n = 0;
   sscanf(time, "%d h%d m%d s %n", &h,&m, &s, &n);
   if (n == 0 || time[n] != '\0') return -1;  // failure
   return h*3600 + m *60 + s;
}

The "%n" saves the count of characters scanned. Since is used at the end of the format, testing to see if it is non-zero (the original value) and that time ended at the null character insures the integrity of the scan.

@chqrlie suggested a mis-match of C# functionality.

   // to match C#
   if (n == 0) return -1;  // failure

Answer 3

A much simpler solution is possible:

int getseconds( const char* time)
{
    int h, m, s = 0 ;
    char sdelim = 0 ;
    int check = sscanf( time, "%dh %dm %d%c", &h, &m, &s, &sdelim ) ;
    if( check == 3 && sdelim = 's' )
    {
        s = ((h * 60) + m) * 60 + s ;
    }

    return s ;
}