I came across this piece of code in C:
#include <stdio.h>
main( )
{
int i = 5;
workover(i);
printf("%d",i);
}
workover(i)
int i;
{
i = i*i;
return(i);
}
I want to know how the declaration of the function "workover" is valid? What happens when we don't mention the return type of a function? (can we return anything?).The parameter is also just a variable name, how does this work?
If you do not specify a return type or parameter type, C will implicitly declare it as int
.
This is a "feature" from the earlier versions of C (C89 and C90), but is generally considered bad practice nowadays. Since the C99 standard (1999) does no longer allow this, a compiler targeting C99 or later will likely give you a warning similar to the following:
program.c: At top level:
program.c:8:1: warning: return type defaults to ‘int’
workover(i)
^
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