warning C4127: conditional expression is constant

Question

#define VALUE_MAX 300
int main() {
   if(VALUE_MAX)
      printf("The value is %d",VALUE_MAX);
   return 0;
}

When I try to compile the above program if(VALUE_MAX) is showing a warning

conditional expression is constant.

How to solve the above warning?

Answer 1

since the if condition becomes always true that is a constant ...

int main()
 {
// if(VALUE_MAX)
#ifdef VALUE_MAX
 printf("The value is %d",VALUE_MAX);
#endif
 return 0;
 }

Answer 2

In your code, VALUE_MAX is not a variable, it's a MACRO. MACROs can be considered as a textual replacement at preprocessing time. So,

  if(VALUE_MAX)

which is translated to

 if (300)

is always TRUE. It is equivalent to

  if(1)

which is having essentially no effect. The code block under the if condition will execute unconditionally.

---

EDIT: ( Elaboration for better understamding )

An if statement is called a selection statement . The syntax of simple if statement is

 if ( expression ) statement

based on the evaluation of expression , it is decided whether the following statement (or block) will be executed.

In case of your code,

  if(VALUE_MAX)

always evaluates to TRUE. In this scenario, the use of if statement is meaningless. YOu can get rid of the if statement altogether.

Answer 3

VALUE_MAX is replace with a number therefore the condition is replaced with

if(300)

which is always true.

ways to get around the warning:

1. change VALUE_MAX to a variable

int VALUE_MAX = 300;

2. change the condition to ifdef

#ifdef VALUE_MAX printf("The value is %d",VALUE_MAX); #endif

Answer 4

You probably want a (pre-)compile time macro "if" ( #ifdef ), rather than a runtime "if" ( if (…) ):

#define VALUE_MAX 300

int main() {
#ifdef VALUE_MAX
   printf("The value is %d", VALUE_MAX);
#endif
   return 0;
}

The code between #ifdef and #endif will be compiled if you have #define d the VALUE_MAX macro.