如何根据 C “字符串”的长度闯入 Xcode 的调试器？

Question

After searching online I found two ways to calculate the length of a C "String" [really character array]

C array length:

*(&arr + 1) - arr

OR

sizeof(arr) / sizeof(arr(0))

only the one using sizeof works in Xcode's debugger.

I need to break into the debugger when a variable declared as "char* output[]" is 5.

I tried "sizeof(output)/sizeof(output[0]) == 5" and it doesn't seem to be working, the break never triggers.

What am I doing wrong?

Answer 1

You have to execute something based on the test, otherwise the compiler will ignore it as a meaningless statement (although it should have flagged it if warnings were enabled).

void r(void)
{   
    printf("Condition was triggered\n");
}

main()
{
    do_stuff();
    if(sizeof(output)/sizeof(output[0]) == 5)
    {
       r();
    }

then set a breakpoint on r().