Thread: Declaring Variable

Ciao,

This is a function to find sum of squared number between [a,b] interval. So my question is how do i get an output if i dont declare acc and bcc. Of course the output is wrong but, shouldnt compiler give an error? Or what is the value of an undeclared variable?

Code:

#include <stdio.h>

int main (void) {
	
	int a,b,ac,acc,bc,bcc;
	int result;
	scanf("%d%d",&a,&b);
	acc=0;
	bcc=0;
	if(a<=100 && b<=100){
		for(ac=1; ac<=a-1; ac++){
			acc+=ac*ac;
		}
		for(bc=1;bc<=b; bc++){
			bcc+=bc*bc;
		}
	}
	
	result=bcc-acc;
	printf("%d",result);
	return 0;
}

Code:

#include <stdio.h>

int main (void) {
	
	int a,b,ac,acc,bc,bcc;
	int result;
	scanf("%d%d",&a,&b);
	if(a<=100 && b<=100){
		for(ac=1; ac<=a-1; ac++){
			acc+=ac*ac;
		}
		for(bc=1;bc<=b; bc++){
			bcc+=bc*bc;
		}
	}
	
	result=bcc-acc;
	printf("%d",result);
	return 0;
}

Something that is not initialised is not the same as something that is not declared.

Hodor is right, but even the uninitialized case is caught by most C compilers, if you just ask for such warnings.

If I save the latter code as example.c, and compile it using

gcc -Wall -Wextra -O2 example.c -o example

GCC helpfully tells me:

Code:

example.c: In function ‘main’:
example.c:7:10: warning: ignoring return value of ‘scanf’, declared with attribute warn_unused_result [-Wunused-result]
     scanf("%d%d",&a,&b);
          ^
example.c:17:11: warning: ‘bcc’ may be used uninitialized in this function [-Wmaybe-uninitialized]
     result=bcc-acc;
           ^
example.c:17:11: warning: ‘acc’ may be used uninitialized in this function [-Wmaybe-uninitialized]

The return value of scanf() is the number of successful conversions (or EOF if read error, or end of input is reached before the first conversion or match). You really should verify it returned 2 (meaning both a and b were read); and if not, complain and exit.