passing float

[bobetko]

Will somebody please tell why this doesn't work.... I am passing param by value:

float x = myfun();
printf("value = %.1f", x);

.........


float myfun(){
    float c = 0.5;
    return (c);
}

I am getting big numbers, something like 803459439.0 etc....
What's going on here? Can I pass float type number by value?

I also tried this printf("value = %.1f", 5); and I got very big number.... several lines of numbers.... Why?

Thanks in advance....

 

[Diancecht]

Try

float c = 0.5f;

Cheers,
Dian

 

[DaveSinkula]

I also tried this printf("value = %.1f", 5); and I got very big number.... several lines of numbers.... Why?

Because 5 is an int. Try this.

printf("value = %.1f", 5[blue].0F[/blue]);

 

[Salem]

> What's going on here? Can I pass float type number by value?
First guess would be you didn't prototype your function.
Some compilers don't check too well, and as written the compiler might assume your function returns int. Post complete programs next time, we can't tell what you've really done.

> printf("value = %.1f\n", 5);
[tt]$ gcc -W -Wall -ansi -pedantic foo.c
foo.c: In function ‘main’:
foo.c:9: warning: format ‘%.1f’ expects type ‘double’, but argument 2 has type ‘int’
[/tt]
From my compiler.

--

 

[ArkM]

There are 2 different moments here.
1. If no prototype before myfunc() call (see Salem's post), int myfunc() assumed then int-to-float promotion executed in the assignment stmt then float-to-double promotion executed in printf arg list. Of course, the results of these conversions are wrong.
2. printf("value = %.1f", 5); is wrong because of only float-to-double promotion (always!) executed in int printf(const char*,...) arg list, int arg passed as is but format specification wants double then treated int as double.
Moral:
1. Avoid func calls w/o prototypes.
2. Avoid using float type, always use double.