Pelles C forum
C language => Beginner questions => Topic started by: joselitosn on April 02, 2007, 04:35:02 PM
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What's wrong with using "short int" in this code?
#include <stdio.h>
#include <conio.h>
#include <math.h>
double hypotenuse( short int cat_a, short int cat_o );
int main( void )
{
short int ca, co;
printf("First cathetus:\n");
scanf("%d", &ca );
printf("Second cathetus:\n");
scanf("%d", &co );
printf("The hypotenuse of the triangle is %.2f .\n", hypotenuse( ca, co ));
getch();
return( 0 );
}
double hypotenuse( short int cat_a, short int cat_o )
{
double hip, x, y;
printf("%d\n", cat_a );
printf("%d\n", cat_o );
x = pow((double) cat_o, 2 );
y = pow((double) cat_a, 2 );
hip = x + y;
return( sqrt( hip ));
}
The printf show that the variable ca is not passed, even if change it's position inside the hypotenuse call.
It's something related to it being 16-bit? Also it didn't work well with another compilers. Ch Interpreter and lcc-win32 worked, but lcc throws an exception at the end.
Of course changing to "int" work, but i just didn't understand why. Also if i declare "short int ca[2]" and use it as arguments, it works without a problem (except lcc which throws an exception again at the end).
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Prepend "%d" type character with "h" argument-type modifier both in scanf() and printf() calls, that is: use the string "%hd".
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Of course changing to "int" work, but i just didn't understand why. Also if i declare "short int ca[2]" and use it as arguments, it works without a problem (except lcc which throws an exception again at the end).
The problem is, that printf does not know the format (size) of the parameters behind the format-string. With %d you say there comes an integer. If you use a compiler where short int and int have a different size, your code will not work.