invalid type argument of unary '*'

Question

I'm a week into an intro programming class, and I'm having trouble with fixing what's supposed to be a relatively simple code. I keep getting an invalid type argument of unary '*' error.

#include <stdio.h>
#define PI 3.14159; 
int main()
{
   float r;
   float area;
   scanf("%f", &r);
   area = PI * r * r;
   printf("Area is %f", area);
   return 0; 
}

Could someone explain this, and how to fix it?

score 10 · Accepted Answer · answered Sep 03 '13 at 21:54

10

#define PI 3.14159; 
                  ^

Drop the semicolon. Leaving it in, the code will expand to:

area = 3.14159; * r * r;

answered Sep 03 '13 at 21:54

cnicutar

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3

Or better yet, stop using #define. – vipw Sep 03 '13 at 21:55
4

@RobertAleksanderYevdokimov Don't `#define PI 3.14159`, in first place. `#include ` and use `M_PI`. Generally: don't reinvent the wheel, use the standard library whenever possible. – Sep 03 '13 at 21:56
1

`M_PI` isn't part of standard C, is it? – Carl Norum Sep 03 '13 at 21:58
Indeed it isn't: http://stackoverflow.com/questions/5007925/using-m-pi-with-c89-standard – Carl Norum Sep 03 '13 at 22:01
1

@CarlNorum **Must not**. Didn't think of that :-) – cnicutar Sep 03 '13 at 22:01
1

@RyanAmos: What's wrong with using a macro where none is needed is that it circumvents the language, opening up weird errors like this. If the OP had instead written `const float PI = 3.14159f;`, this question wouldn't exist. – Chuck Sep 03 '13 at 23:03

score 2 · Answer 2 · answered Sep 03 '13 at 21:55

2

You have to remove the extra ; in the definition of the macro PI. It is unnecessary for macro, and in your case results in syntax error after expansion.

answered Sep 03 '13 at 21:55

rems4e

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invalid type argument of unary '*'

2 Answers2