what does foo() = "test"; do?-CodePudding

While trying to explore how the compiler handles strings in C I tried the following program that compiles and runs w/o errors, but I cannot understand if it changes any memory locations or has any effect on the program flow. It doesn't seem to do anything.

#include <iostream>
using namespace std;

string str = "this is public string\n";

string foo() {
    return str;
}

int main()
{
    foo() = "test";       //  what does this line do
    cout << str << endl;

    char c;
    cin>>c;
    return 0;
}

It does generate some assembly instructions but I have no idea what they do.

getString() = "test";       //  what does this line do
00007FF68B0D6F1B  lea         rcx,[rbp 0E8h]  
00007FF68B0D6F22  call        getString (07FF68B0D1640h)  
00007FF68B0D6F27  mov         qword ptr [rbp 118h],rax  
00007FF68B0D6F2E  mov         rax,qword ptr [rbp 118h]  
00007FF68B0D6F35  mov         qword ptr [rbp 120h],rax  
00007FF68B0D6F3C  lea         rdx,[string "test" (07FF68B0E1584h)]  
00007FF68B0D6F43  mov         rcx,qword ptr [rbp 120h]  
00007FF68B0D6F4A  call std::basic_string<char,std::char_traits<char>,std::allocator<char> >::operator= (07FF68B0D1037h)  
00007FF68B0D6F4F  nop  
00007FF68B0D6F50  lea         rcx,[rbp 0E8h]  
00007FF68B0D6F57  call  std::basic_string<char,std::char_traits<char>,std::allocator<char> >::~basic_string<char,std::char_traits<char>,std::allocator<char> > (07FF68B0D1140h) 
cout << str << endl;

CodePudding user response：

It, essentially, doesn't do anything useful.

The foo function returns a copy of the string str, and you assign to this copy. It's similar to:

{
    string temp;
    temp = foo();
    temp = "test";

    // Here the temp object is destructed, and the modifications
    // made to it are lost
}

Because the statement doesn't have any observable effect, the as-if rule allows the compiler to skip not only the assignment but also the call to foo itself.

Things would be very different if the foo function returned a reference to the str object:

string& foo() {
    return str;
}

Because then the assignment would assign to the actual str object instead:

{
    string& temp_ref = foo();
    temp_ref = "test";  // Really assigns to str
}

CodePudding user response：

foo() returns a copy of str.

foo() = "test";

This line changes the copy's contents to "test" but does not store it anywhere, so effectively it does nothing.