strtol在调试和发布模式上有所不同

Question

I use following code in c++ compiling with visual studio 2008:

char input = 'K';
long output= strtol(&input , NULL, 36);

In debug mode it works fine, the output is 20.
But in release mode it makes strange outputs, like 604663109.
It also works, when I set disable code optimization, but that can't be the solution.
I also know, how to calculate it by hand, but I want to know, why there is a difference between debug and release mode?

EDITED: Sorry, it has to be char and not char*. But same error.

Answer 1

strtoul doesn't work on char, only on null terminated strings. You'll have to use a null terminated string, or find another way. Eg

char input = 'K';
...
char temp[2] = { input, '\0' };
long output = strtol(temp, NULL, 36);

Answer 2

Since &input doesn't point to a null-terminated character array, the code has undefined behaviour.

This should work:

const char* input = "K";
long output = strtol(input , NULL, 36);

The reason it appears to work in the debug build is probably that the compiler has inserted "magic" values around the input variable in order to enable detection of stack corruption, and a zero happens to be placed at &input + 1 .

Answer 3

Your call is wrong, you shouldn't take the address of the pointer!

It should be:

strtoul(input, NULL, 36);