条件运算和花括号会影响代码吗？

Question

I am wondering if a conditional operator could actually prevent other unrelated code from working. For example below:

typedef char WCHAR_T;
#define STRLEN(x) strlen(x)
if (argc > 2)
{
    WCHAR *pFileName = argv[1];
    basic_string <WCHAR> strFileName(pFileName, STRLEN(pFileName));
}

In the code above, pFileName, argv[1] and strFileName have nothing to do with the comparison argc > 2 . Assuming that the command line arguments are perfectly fine. However, this code would not work with that comparison. Also, even if I change the code to the following format:

typedef char WCHAR_T;
#define STRLEN(x) strlen(x)

{
    WCHAR *pFileName = argv[1];
    basic_string <WCHAR> strFileName(pFileName, STRLEN(pFileName));
}

Still doesn't work. EDIT: By "doesn't work", I mean compiler gives error message such as "strFileName" was not declared, which means this declaration didn't run at all.

I have no idea why the curly braces would have such big impact at the code. When I get rid of the curly braces, the code works like magic...Could anyone explain this please? Thanks.

Answer 1

Because strFileName only exists within the braces. Braces define the scope, local variables are only known within the scope they're defined in.

Answer 2

i guess you're try to use the variable strFileName from somewhere outside the braces, which is not possible.

the { and } braces define a block and variables declared inside are only existent inside this block. a workaround would be to declare strFileName outside of the block and assign it's value from inside

typedef char WCHAR_T;
#define STRLEN(x) strlen(x)

basic_string <WCHAR> strFileName;

if (argc > 2)
{
    WCHAR *pFileName = argv[1];
    strFileName.assign(pFileName, STRLEN(pFileName));
}

std::cout << strFileName << std::endl; // should work now

Answer 3

I see a define for WCHAR_T, but what's used in the code is WCHAR. Is WCHAR defined anywhere?