嵌套for循环的说明

Question

嗨,有人可以请你解释下面的循环如何打印45.我想我不太明白这些嵌套循环是如何工作的:

#include <stdio.h>

int main() {
    int x, y, n;
    n = 0;
    for (x = 0; x < 10; x++)
        for (y = 0; y < x; y++)
            n++;

    printf("%d\n", n);
}

Answer 1

尝试扩展第一个循环,帮助自己理解.如果你这样做,你会得到这样的代码:

#include <stdio.h>  
int main() { 
    int x, y, n;
    n = 0;

    /* x=0 */
    for (y=0; y<0; y++)
        n++; /* doesn't get hit n=0 */

    /* x=1 */
    for (y=0; y<1; y++)
        n++; /* gets hit once n=1 */

    /* x=2 */
    for (y=0; y<2; y++)
        n++; /* gets hit twice n=3 */

    /* x=3 */
    for (y=0; y<3; y++)
        n++; /* gets hit three times n=6 */

    /* x=4 */
    for (y=0; y<4; y++)
        n++; /* gets hit four times n=10 */

    /* x=5 */
    for (y=0; y<5; y++)
        n++; /* gets hit five times n=15 */

    /* x=6 */
    for (y=0; y<6; y++)
        n++; /* gets hit six times n=21 */

    /* x=7 */
    for (y=0; y<7; y++)
        n++; /* gets hit seven times n=28 */

    /* x=8 */
    for (y=0; y<8; y++)
        n++; /* gets hit eight times n=36 */

    /* x=9 */
    for (y=0; y<9; y++)
        n++; /* gets hit nine times n=45 */

    /* x=10 exit loop */

    printf ("%d\n", n); /* prints 45 */
}