如何在C中将一个字符串分为3个短裤，其余部分分为一个字符串（此字符串为空格）？

Question

How can I separate a string that has white spaces between the 3 shorts and between the rest of the string to 4 different strings.

Example:

"123 402 10 aaa bbb cc".

What I want is simply

 short i=123;
 short j=402;
 short y=10;
 char * c="aaa bbb cc".

I was trying to use sscanf to do it but I can't seem to get the hang of getting that last string to work cause of the white space.

Answer 1

One way to do it could be:

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

int main()
{
    const char* str = "123 402 10 aaa bbb cc";
    short i,j,y;
    char c[256];
    if (sscanf(str, "%hd %hd %hd %255[^\n]", &i, &j, &y, c) == 4) {
        printf("i=%hd j=%hd y=%hd c=\"%s\"\n", i, j, y, c);
    }
    return 0;
}

Answer 2

It's not that difficult:

#include <stdio.h>

int main() {
  short i, j, y;
  char text[80];

  if (sscanf("123 402 10 aaa bbb cc\nsecond line", "%hd %hd %hd %79[^\n]", &i, &j, &y, text) == 4) {
    printf("success: i=%d, j=%d, y=%d, text=%s\n", i, j, y, text);
  }
  return 0;
}

Note that you have to allocate the buffer for the string yourself and make sure that no buffer overflow happens.

Answer 3

You don't need sscanf . You can use strchr to find the spaces, atoi to turn strings into integers, and simple assignment to turn the spaces into terminating zeroes.

You can also do this:

#include <stdio.h>
int main(void)
{
 const char *test="123 402 10 aaa bbb cc";
 short i, j, y;
 char c[128];
 sscanf(test, "%hd%hd%hd %[^\n]s", &i, &j, &y, c);
 printf("i=%d j=%d y=%d c='%s'\n", i, j, y, c);
}

Yields: i=123 j=402 y=10 c='aaa bbb cc'