使用 cout API 时出现乱码 output

Question

I am trying to run this simple code in VS 2015

#include "stdafx.h"
# include <iostream>

int main()
{
    char * szOldPath = "\"C:\icm\scripts\StartupSync\runall.bat\" nonprod"; 
    std::cout << szOldPath << std::endl;

    return 0;
}

However, the output of szOldPath is not proper and the console is printing--

unall.bat" nonprodupSync

I suspect this might be because of Unicode and I should be using wcout. So I disabled Unicode by going to Configuration Properties -> General --> Character Set and tried setting it to Not Set or Multi Byte. But still running into this issue.

I understand it is not good to disable UNICODE but I am trying to understand some legacy code written in our company and this experiment is a part of this exercise,is there any way I can get the cout command to print szOldPath successfully?

Answer 1

Your issue has nothing to do with Unicode.

\r is the escape sequence for a carriage return. So, you are printing out "C:\icm\scripts\StartupSync , and then \r tells the terminal to move the cursor back to the beginning of the current line, and then unall.bat" nonprod is printed, overwriting what was already there.

You need to escape all of the \ characters in your string literal, just like you had to escape the " characters.

Also, your variable needs to be declared as a pointer to const char when assigning a string literal to the pointer. This is enforced in C++11 and later:

#include "stdafx.h"
#include <iostream>

int main()
{
    const char * szOldPath = "\"C:\\icm\\scripts\\StartupSync\\runall.bat\" nonprod"; 
    std::cout << szOldPath << std::endl;

    return 0;
}

Alternatively, in C++11 and later, you can use a raw string literal instead to avoid having to escape any characters with a leading \ :

const char * szOldPath = R"("C:\icm\scripts\StartupSync\runall.bat" nonprod)";