它将执行2次

Question

我是C编程新手。我遇到了for循环示例。我不了解循环的某些部分。输出是8。直到4为止，我都不知道b是怎么回事。这是我的代码。

int a = 4;
int b = 2;
int result = 0;

for(int count = 0; count != b; count++) {
    result = result + a;
}

printf("a times b is %i\n", result);

return 0;

Answer 1

有时候，最简单的方法是让程序进行自我解释：

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

int main(void){
    int a = 4;
    int b = 2;
    int result = 0;

    for(int count = 0; count != b; count++) {
        printf("a = %3d, b = %3d, count = %3d, result = %3d\n", a, b, count, result);
        result = result + a;
        printf("a = %3d, b = %3d, count = %3d, result = %3d\n", a, b, count, result);
    }
    printf("a = %3d, b = %3d\n", a, b);

    printf("a times b is %i\n", result);

    return 0;
}

输出

a =   4, b =   2, count =   0, result =   0
a =   4, b =   2, count =   0, result =   4
a =   4, b =   2, count =   1, result =   4
a =   4, b =   2, count =   1, result =   8
a =   4, b =   2
a times b is 8

如您所见，b不变。 count等于count时，b会发生变化并退出循环。

Answer 2

变量b不递增，变量count从0开始并在for loop中递增。当count变量变为2时，循环结束。因此，循环运行两次（计数0和计数1），结果为4 + 4 = 8。

Answer 3

首先，count = 0与b = 2不同，因此循环开始。这样，循环的第一次迭代就会得出结果，

result = 0 + 4

由于循环结束，count变量应继续count++，这意味着count = count + 1。因此，count = 0 + 1 = 1的价值与b = 2不再相同。

再次进入循环，

result = 4  + 4 # where the first number came from the result of first loop

现在，count++生成的count = 2现在与b = 2的值相同。然后，循环条件不匹配count != 2，不再循环并显示值result = 8。

Answer 4

它将执行2次

count = 0 ==> result = 0 + 4

并计数= 1 ==> result = 4 + 4 ==> result = 8

当计数达到2时2！= 2部分将为假，我们将退出for循环

Answer 5

该循环只运行两次，我们可以像这样干循环：变量的初始值：-> a = 4，b = 2，计数= 0，结果= 0

loop first run -> 
count = 0
count != 2 -> that is ->  0 != 2  => true
result = result + a   ->  0 + 4  = 4

second run ->
count = 1 
count != b -> that is ->  1 != 2  => true
result = result + a   ->  4 + 4 = 8

third run ->
count = 2
count != b -> that is ->  2 != 2 => false 
stop loop. 
And the final resultent values are as follow: a= 4, b= 2 , count = 2 and result = 8

此for循环如何工作？

5 个答案:

它将执行2次