C宏，带有表达式不必要的结果 [英] C macro with expression unwanted result

问题描述

我正在运行以下程序并获得9 7的结果，我理解为什么9是输出，但是我不知道为什么我得到7作为输出.

I am running the following program and getting a result as 9 7, I understood why 9 is the output but I can't figure out why I'm getting 7 as output.

#include<stdio.h>
#define sqr(i) (i*i)

int main()
{
    printf("%d %d", sqr(3), sqr(3+1));
    return 0;
}

第二个功能是sqrt(3+1)微型如何扩展以及我如何获得7输出?

For the second function that is sqrt(3+1) how the micro is getting expanded and how Im getting 7 output?

推荐答案

您可以让编译器或IDE对文件进行预处理，并向您展示宏的扩展方式.

You can have the compiler or IDE preprocess the file and show you how the macro expanded.

在您的情况下，sqr(3+1)扩展为(3+1*3+1).现在，C运算符的优先级意味着乘法在加法之前完成.所以(3+1*3+1)-> (3+3+1)-> (7).

In your case sqr(3+1) expands to (3+1*3+1). Now the precedence of C operators means that the multiplication is done before the addition. So (3+1*3+1) -> (3+3+1) -> (7).

您可以通过以下方式定义宏来解决此问题，并在参数周围加上括号:

You can fix this by defining your macro this way, with parentheses around the argument:

#define sqr(i) ((i)*(i))

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