可以得到的printf在C程序会自动将取代？ [英] Can printf get replaced by puts automatically in a C program?

问题描述

#include <stdio.h>

int puts(const char* str)
{
    return printf("Hiya!\n");
}

int main()
{
    printf("Hello world.\n");
    return 0;
}

这code输出你好！当运行。可能有人解释为什么？

This code outputs "Hiya!" when run. Could someone explain why?

编译行是：
GCC的main.c

编辑：它现在是纯C，任何多余的东西已经从编译行删除

it's now pure C, and any extraneous stuff has been removed from the compile line.

推荐答案

是，编译器可通过替换为的printf 调用的相当于的来电看跌。

Yes, a compiler may replace a call to printf by an equivalent call to puts.

由于您定义自己的函数看跌具有相同名称作为标准库函数，你的程序的行为是不确定的。

Because you defined your own function puts with the same name as a standard library function, your program's behavior is undefined.

参考： N1570 7.1.3：

在任一下列条款与外部链接的所有标识符[这包括看跌]总是保留用作与外部连接的标识符。结果
   ... 结果
  如果程序声明或在一个限定的标识符
  上下文中它被保留（比其它由7.1.4所允许），或限定一个保留
  标识为宏名，行为是不确定的。

All identifiers with external linkage in any of the following subclauses [this includes puts] are always reserved for use as identifiers with external linkage.
...
If the program declares or defines an identifier in a context in which it is reserved (other than as allowed by 7.1.4), or defines a reserved identifier as a macro name, the behavior is undefined.

如果您删除自己的看跌函数，检查装配上市，你的可能的找到使通话在生成的code，你在源$ C ​​$ C名为的printf 。 （我见过的gcc执行这个特别的优化。）

If you remove your own puts function and examine an assembly listing, you might find a call to puts in the generated code where you called printf in the source code. (I've seen gcc perform this particular optimization.)

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