Evaluate the following statements, and show the value that will be in the receiv

Question

Evaluate the following statements, and show the value that will be in the receiving variable. (Hint: Do these by hand (no calculator), and then check / correct them in a program. The tests and exams for this course will not allow calculators or computers.)

long int ivar1, ivar2 = 20, ivar3 = 6;
double dvar1, dvar2 = 1.2, dvar3 = 4.5;

ivar1 = dvar2; __
ivar1 = ivar2 / ivar3; ____
ivar1 = ivar2 % ivar3 * 2; ____
ivar1 = ivar2 / ivar3 % 2; ____
ivar1 = ivar2 + dvar2 * ivar2; ____
ivar1 = (dvar2 * 2) + (ivar2 / 19); ____

dvar1 = ivar3 * dvar2; ____
dvar1 = (int) dvar2; ____
dvar1 = (double) ivar2 / ivar3; ____
dvar1 = dvar2 * dvar3 + (ivar2 % 3); ____

ivar2 *= ivar3 + 14; ____
ivar3 += dvar2; ____

dvar2 /= 1.0 * dvar3; ____
dvar3 += ivar3++; ____

How do I run it in a program to see if my answers are correct?

Explanation / Answer

#include <stdio.h>

int main( int argc, char** argv ) {

long int ivar1, ivar2 = 20, ivar3 = 6;
double dvar1, dvar2 = 1.2, dvar3 = 4.5;

ivar1 = dvar2;

printf( "ivar1 = %d ", ivar1 );

ivar1 = ivar2 / ivar3;

printf( "ivar1 = %d ", ivar1 );

ivar1 = ivar2 % ivar3 * 2;

printf( "ivar1 = %d ", ivar1 );

/*

A few ivar's omitted for brevity's sake
*/

dvar1 = ivar3 * dvar2;

printf( "dvar1 = %f ", dvar1 );

dvar1 = (int) dvar2;

printf( "dvar1 = %f ", dvar1 );

}

In short, printf() is your friend -- adding an extra printf() or two here or there will show you exactly what is going on in your program, and it's sometimes much easier than using a debugger.

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