说我有char* b = "2 3, 32 3, 6 8, 9 10"
和方法randMethod(int x, int y)
我将如何继续处理将每2个int作为randMethod(x,y)的输入的字符串?
因此,最终结果如下:
randMethod(2, 3);
randMethod(32, 3);
randMethod(6, 8);
randMethod(9, 10);
最佳答案
像这样:
#include <stdio.h>
#include <string.h>
void randMethod(int x, int y)
{
printf("%d %d\n", x, y);
}
int main()
{
const char * b = "2 3, 32 3, 6 8, 9 10";
int x, y;
while (sscanf(b, "%d %d", &x, &y) == 2) {
randMethod(x, y);
b = strchr(b, ',');
if (b == NULL)
break;
b += 1;
}
return 0;
}
编译与执行:
pi@raspberrypi:/tmp $ gcc -pedantic -Wextra c.c
pi@raspberrypi:/tmp $ ./a.out
2 3
32 3
6 8
9 10
pi@raspberrypi:/tmp $