这段 C 代码中的括号如何使结果如此不同？
这段代码的输出：
#define RECIPROCAL_1(x) 1/(x)
#define RECIPROCAL_2(x) 1/x
main()
{
float x=8.0, y;
y = RECIPROCAL_1(x+10.0);
printf("1/%3.1f = %8.5f\n", x, y);
y = RECIPROCAL_2(x+10.0);
printf("1/%3.1f = %8.5f\n", x, y);
}
是输出=
1/8.0 = 0.05556
1/8.0 = 10.12500
我看不出它是如何工作的。我很感激任何提示。
The output for this code:
#define RECIPROCAL_1(x) 1/(x)
#define RECIPROCAL_2(x) 1/x
main()
{
float x=8.0, y;
y = RECIPROCAL_1(x+10.0);
printf("1/%3.1f = %8.5f\n", x, y);
y = RECIPROCAL_2(x+10.0);
printf("1/%3.1f = %8.5f\n", x, y);
}
is output =
1/8.0 = 0.05556
1/8.0 = 10.12500
I can't see how this works though. I appreciate any tips .
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宏替换的扩展如下：
becomes
和
becomes
因为
/的优先级高于+，所以y的值不同。这是一个很好的例子，说明了为什么挑剔的程序员在没有其他解决方案可行的情况下只使用宏。即便如此，那些觉得有必要使用宏的挑剔的程序员总是会使用足够的括号来确保避免此类陷阱。
The macro substitutions are expanded like this:
becomes
and
becomes
Because
/has a higher precedence that+the values foryare different.This is an excellent example of why the discerning programmer only reaches for macros when no other solution is viable. And even then, those discerning programmers that feel compelled to use macros, will always use sufficient parenthesise as to make sure such pitfalls are avoided.
宏只是做非常基本的标记替换，所以
相当于
Macros just do very basic token substitution, so
is equivalent to
让我们展开宏：
与 C 中运算符优先级的方式相比
，除数在加法之前执行，使得 #2 (1/x) + 10 而不是 1 / (x + 10)。这就是为什么你应该总是给你的宏加上括号。
Lets expand the macros:
versus
The way operator precedence is done in C, the divisor is performed before the addition is, making #2 (1/x) + 10 instead of 1 / (x + 10). That is why you should always parenthesize your macros.