Is the following C code safe?

Question

#include<cstdio>
#include<stdlib.h>

int main()
{
    char* ptr=NULL;
    printf("%s",ptr);
    return 0;
}

It prints (null) as output. The above is a sample code. In real code i get char* as a return of a function and i wish to print the character string for logging. However, NULL is also a valid return value of that function and so i am wondering if a null check is required before printing the character string?

char* ptr=someFuncion();
// do i need the following if statement?
if(ptr!=NULL)
{
  printf("%s",ptr);
}

I just want to be sure that the output would be same ie if ptr=NULL then output should be (null) on all platforms and compilers and the above code(without if statement) would not crash on any C standard compatible platform.

In short, is the above code(without the if statement) standard compatible?

Thanks for your help and patience :)

Regards

lali

Answer 1

In short, is the above code(without the if statement) standard compatible?

No. ISO/IEC 9899:1999 (the C standard document) makes no statement about what should happen if ptr is NULL, so the behaviour is undefined . The library you used merely was friendly enough to give you some helpful output ("(null)") instead of crashing.

Include the explicit check for NULL.

Answer 2

Do you mean something like this?

char* result = foo ();
  printf ("result is %s\n", (result ? result : "NULL"));

Answer 3

如果有疑问，你不应该依赖于实现细节并执行额外的(ptr != NULL) - 这也是很好的编码实践。

Answer 4

根据我的经验，通常你会在没有if语句的情况下处于清晰状态，尽管我倾向于避免做出你所说的习惯......这已经很长一段时间了，但IIRC是我曾经使用过的Sun编译器如果你将一个NULL char *传递给printf（）会导致部分或全部崩溃，所以包含检查更简单，更安全......我打算在使用宏形式的注释中，但我发现自从我开始打字以来，我在30秒内被其他3人殴打过它:)