What is the effect of this statement in C program define PI 3.14159?

Answer 1

All pre-processor directives begin with a # symbol. One of the most-used pre-processor directives is the #define directive, which has the following syntax:

#define SYMBOL definition

This defines a macro. During preprocessing, all occurrences of SYMBOL within your source code will be replaced with whatever is written in the definition (which includes everything up to the end of the line).

#define PI 3.14159

Here, all occurrences of the symbol PI within your source code will be replaced with the character sequence 3.14159. So if your source contained the following function:

double area_of_circle (double radius) {

return 2*PI*radius*radius; // 2 PI r squared

}

The compiler will see the following instead:

double area_of_circle (double radius) { return 2*3.14159*radius*radius;

}

While this may well seem a convenient method of defining constants, it is not. Macros should never be used to define constants. If you need a constant, use an actual constant. If the constant must be calculated at compile time, then use a constant expression. In this case we can define PI as follows:

constexpr double PI (void) {

return 4.0 * atan (1.0);

}

Note that the literal value 3.14159 takes no account of the implementation's precision because the compiler will convert it to a value of 3.141590. By defining the constant expression, the compiler will use a value of 3.14159265359..., including as many digits of precision as the implementation will physically allow, and thus minimising rounding errors.

Macros (#defines) should only be used for conditional compilation, never to define constants.