printf输出错误 [英] printf giving wrong output

问题描述

#include<stdio.h>
#define square(x) x*x

    void main()
    {
        int i;
        i = 8 / square(4);
        printf("%d %d", i, 8/square(4));
    }

给出的输出:8 8

但是如果我写下面的代码:

but if I write below code :

#include<stdio.h>
#define square(x) x*x

    void main()
    {
        float i;
        i = 8 / square(4);
        printf("%f %f", i, 8/square(4));
    }

给出的输出:8.000000 0.000000

Gives Output : 8.000000 0.000000

为什么这样???请解释

Why like that??? please explain

推荐答案

问题不仅与格式说明符有关，而且还与您定义宏的方式有关.应该是:

The problems are not just with the format specifier but also the way you have defined your macro. It should be:

#define square(x) ((x)*(x))

宏也不是安全类型.现在，如果您对结果进行转换，您将看到正在发生的事情，因为4的平方是16，而8/16的平方是0.5，被截断为int，因此变为0.对于正确的值，这是您应该键入的方式:

Also macros are not type safe. Now if you cast your results you will see what is happening, since the square of 4 is 16 and 8/16 is 0.5 which gets truncated to int hence becomes 0. For proper values this is how you should typecast:

printf("%d %d", (int)i, (int)(8/square(4)));
printf("\n%f %f", (float)i, (float)8/((float)square(4)));

示例输出:

0 0
0.000000 0.500000

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