编译器将printf更改为puts [英] Compiler changes printf to puts

问题描述

考虑以下代码:

#include <stdio.h>

void foo() {
    printf("Hello world\n");
}

void bar() {
    printf("Hello world");
}

这两个函数生成的程序集是:

The assembly produced by both these two functions is:

.LC0:
        .string "Hello world"
foo():
        mov     edi, OFFSET FLAT:.LC0
        jmp     puts
bar():
        mov     edi, OFFSET FLAT:.LC0
        xor     eax, eax
        jmp     printf

现在，我知道了 puts与printf之间的区别，但是我发现这很有趣，因为gcc能够自省const char *并弄清楚是调用printf还是puts.

Now I know the difference between puts and printf, but I find this quite interesting that gcc is able to introspect the const char* and figure out whether to call printf or puts.

另一个有趣的事情是，在bar中，编译器将返回寄存器(eax)归零，即使它是void函数.为什么要在那做而不是在foo做?

Another interesting thing is that in bar, compiler zero'ed out the return register (eax) even though it is a void function. Why did it do that there and not in foo?

我是否假设编译器自检了我的字符串"是正确的，还是对此有另一种解释?

Am I correct in assuming that compiler 'introspected my string', or there is another explanation of this?

推荐答案

我是否假设编译器自检了我的字符串"是正确的，还是对此有另一种解释?

Am I correct in assuming that compiler 'introspected my string', or there is another explanation of this?

是的，这正是发生的情况.这是编译器完成的非常简单且常见的优化.

Yes, this is exactly what happens. It's a pretty simple and common optimization done by the compiler.

由于您的第一个printf()通话是:

Since your first printf() call is just:

printf("Hello world\n");

等效于:

puts("Hello world");

由于puts()不需要扫描和解析用于格式说明符的字符串，因此它比printf()快得多.编译器会注意到您的字符串以换行符结尾，并且不包含格式说明符，因此会自动转换该调用.

Since puts() does not need to scan and parse the string for format specifiers, it's quite faster than printf(). The compiler notices that your string ends with a newline and does not contain format specifiers, and therefore automatically converts the call.

这还节省了一些空间，因为现在只需要在生成的二进制文件中存储一个字符串"Hello world".

This also saves a bit of space, since now only one string "Hello world" needs to be stored in the resulting binary.

请注意，通常对于以下形式的调用是不可能的:

Note that this is not possible in general for calls of the form:

printf(some_var);

如果some_var不是简单的常量字符串，则编译器无法知道它是否以\n结尾.

If some_var is not a simple constant string, the compiler cannot know if it ends in \n.

其他常见的优化方法是:

Other common optimizations are:

• strlen("constant string")可能会在编译时求值并转换为数字.
如果编译器确定location1和location2不重叠，则
• memmove(location1, location2, sz)可能会转换为memcpy().
较小的
• memcpy()可以在一条mov指令中转换，即使大小较大，有时也可以内联以更快地调用.

• strlen("constant string") might get evaluated at compile time and converted into a number.
• memmove(location1, location2, sz) might get transformed into memcpy() if the compiler is sure that location1 and location2 don't overlap.
• memcpy() of small sizes can be converted in a single mov instruction, and even if the size is larger the call can sometimes be inlined to be faster.

另一个有趣的事情是，在bar中，编译器将返回寄存器(eax)归零，即使它是void函数.为什么要在那做而不是在foo做?

Another interesting thing is that in bar, compiler zero'ed out the return register (eax) even though it is a void function. Why did it do that there and not in foo?

请参阅此处:为什么在通话前将％eax归零来打印吗?

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