相关疑难解决方法(0)

void main() {
    if("a" == "a")
      printf("Yes, equal");  
    else
      printf("No, not equal");
}

为什么输出No, not equal？

刚刚在gdb中检查以下内容:

char *a[] = {"one","two","three","four"};
char *b[] = {"one","two","three","four"};
char *c[] = {"two","three","four","five"};
char *d[] = {"one","three","four","six"};

我得到以下内容:

(gdb) p a
$17 = {0x80961a4 "one", 0x80961a8 "two", 0x80961ac "three", 0x80961b2 "four"}
(gdb) p b
$18 = {0x80961a4 "one", 0x80961a8 "two", 0x80961ac "three", 0x80961b2 "four"}
(gdb) p c
$19 = {0x80961a8 "two", 0x80961ac "three", 0x80961b2 "four", 0x80961b7 "five"}
(gdb) p d
$20 = {0x80961a4 "one", 0x80961ac "three", 0x80961b2 "four", 0x80961bc "six"}

我真的很惊讶字符串指针对于相同的单词是相同的.我原以为每个字符串都会在堆栈上分配自己的内存,无论它是否与另一个数组中的字符串相同.

这是某种编译器优化的示例,还是这种字符串声明的标准行为？

看看这段代码:

#include <iostream>
using namespace std;

int main()
{
    const char* str0 = "Watchmen";
    const char* str1 = "Watchmen";
    char* str2 = "Watchmen";
    char* str3 = "Watchmen";

    cerr << static_cast<void*>( const_cast<char*>( str0 ) ) << endl;
    cerr << static_cast<void*>( const_cast<char*>( str1 ) ) << endl;
    cerr << static_cast<void*>( str2 ) << endl;
    cerr << static_cast<void*>( str3 ) << endl;

    return 0;
}

这会产生如下输出:

0x443000
0x443000
0x443000
0x443000

这是在Cygwin下运行的g ++编译器.即使没有打开优化(),指针也指向相同的位置.-O0

编译器是否总是进行优化以至于它会搜索所有字符串常量以查看它们是否相等？可以依赖这种行为吗？