#include<stdio.h>
void function(int);
int main()
{
int x;
printf("Enter x:");
scanf("%d", &x);
function(x);
return 0;
}
void function(int x)
{
float fx;
fx=10/x;
if(10 is divided by zero)// I dont know what to put here please help
printf("division by zero is not allowed");
else
printf("f(x) is: %.5f",fx);
}
views:
258answers:
4
+5
A:
This should do it. You need to check for division by zero before performing the division.
void function(int x)
{
float fx;
if(x == 0) {
printf("division by zero is not allowed");
} else {
fx = 10/x;
printf("f(x) is: %.5f",fx);
}
}
Eric
2010-03-21 01:58:48
+2
A:
#include<stdio.h>
void function(int);
int main()
{
int x;
printf("Enter x:");
scanf("%d", &x);
function(x);
return 0;
}
void function(int x)
{
float fx;
if(x==0) // Simple!
printf("division by zero is not allowed");
else
fx=10/x;
printf("f(x) is: %.5f",fx);
}
Carlucho
2010-03-21 01:59:26
That's only for floating point exceptions though ? The example above is for an integer divide by zero.
Paul R
2010-03-21 10:18:29
A:
By default in UNIX, floating-point division by zero does not stop the program with an exception. Instead, it produces a result which is infinity
or NaN
. You can check that neither of these happened using isfinite
.
x = y / z; // assuming y or z is floating-point
if ( ! isfinite( x ) ) cerr << "invalid result from division" << endl;
Alternately, you can check that the divisor isn't zero:
if ( z == 0 || ! isfinite( z ) ) cerr << "invalid divisor to division" << endl;
x = y / z;
Potatoswatter
2010-03-21 02:05:05
It's not a floating point divide by zero though - it's integer (the result of the integer division expression is cast to a float afterwards).
Paul R
2010-03-21 10:20:49
@Paul: That is true in his code, but I didn't replicate his code. I added a comment, does that help?
Potatoswatter
2010-03-21 11:29:24