Chapter5-appendix1

Chapter 5 – Appendix

5/18/2008

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1) Write a program that produces this output: 0 1 2 3 4 5 6

1 2 4 8 16 32 64

Or double

Note: The prototype of pow() function is: double pow(double, int) or float pow(float, int) Solution 1: Using pow() function. Using for #include<stdio.h> #include<math.h> int main(void) { int i; double x=2.0,y; for(i=0; i<7; i++) { //printf("%d %3.0f\n",i,pow(x,i)); y = pow(x,i); printf("%d %3.0f\n",i,y); } return(0); }

Solution 2: without using the function pow(). Using while

Using for

#include<stdio.h>

#include<stdio.h>

int main(void) { int i,product; i=0; product=1;

int main(void) { int i, product=1; for(i=0; i<=6; ++i) { printf("%d%3d\n",i,product); product*=2; }

while(i<=6) { printf("%d %3d\n",i,product); i=i+1; product *=2; } return (0); }

return(0); }

Chapter 5 – Appendix

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2) The program uses loop structure to sum and count all even integers from 2 to 100. Solution: Using for

Using while

Using do-while

#include<stdio.h>

#include<stdio.h>

#include<stdio.h>

int main(void) { int ev, sum=0, count=0;

int main(void) { int ev,sum,count; ev=2; sum=0; count=0;

int main(void) { int ev,sum,count; ev=2; sum=0; count=0;

for(ev=2; ev<=100; ev += 2) { sum += ev; count++; } printf("%d numbers were added.\n",count); printf("Sum of the even numbers is %d\n", sum);

do {

while(ev<=100) { sum = sum + ev; ++count; ev = ev+2; }

sum += ev; ++count; ev = ev+2; } while(ev<=100);

printf("%d numbers were added.\n",count);

printf("%d numbers were added.\n",count);

printf("Sum of the even numbers is %d\n", sum);

printf("Sum of the even numbers is %d\n", sum);

return(0); }

return(0);

return(0); }

Output:

}

Chapter 5 – Appendix

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3) Write a program that calculate grade average for 10 students ? Note: average = (double) total/10.0; - if total is declated as int, you need to use double total to convert the value to real. - Another solution is to declare total as double. Solution: Using for

Using while

Using do-while

#include<stdio.h>

#include<stdio.h>

#include<stdio.h>

int main(void) { int counter, total, grade; double average; total=0;

int main(void) { int counter, total, grade; double average; total=0; counter=1;

int main(void) { int counter, total, grade; double average; total=0; counter=1;

while(counter <= 10) { printf("Enter grade %2d: ", counter); scanf("%d",&grade); total += grade; counter++; }

do { printf("Enter grade %2d: ", counter); scanf("%d",&grade); total += grade; counter++; } while(counter <= 10);

average = (double) total/10; printf("Class average is %.2f\n", average);

average = (double) total/10; printf("Class average is %.2f\n", average);

for(counter=1; counter<=10; counter += 1) { printf("Enter grade %2d: ", counter); scanf("%d",&grade); total += grade; } average = (double) total/10; printf("Class average is %.2f\n", average); return(0); }

return(0); }

Output:

return(0); }

Chapter 5 – Appendix

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4) Write a program that calculate grade average until entering -1 to exit the loop ? Note 1: Do not use for if you use scanf to update the loop control variable. Note 2: If you use the loop as a counter, it will be much better to use for and not while. Solution: Using while #include<stdio.h> int main(void) { int counter, total, grade; double average; total=0; counter=0;

Using do-while #include<stdio.h> int main(void) { int total; // sum of grades int counter; // number of grades entered int grade; // one grade double average;

// don't know number of grades printf("Enter grade, -1 to end: "); scanf("%d",&grade);

/* initialization */ total=0; counter=0;

//stop the loop if grade=-1 while(grade != -1) { total += grade; counter++; printf("Enter grade, -1 to end: "); scanf("%d",&grade); }

// don'n know number of grades do { printf("Enter grade, -1 to end: "); scanf("%d",&grade); if(grade == -1) break; total += grade; counter++; } while(grade != -1); //stop the loop if grade=-1

if(counter != 0 ) { average = (double) total/counter; printf("counter = %d\n",counter); printf("Class average is %.2f\n", average); } else printf("No grades were entered\n");

if(counter != 0) { average = (double) total/counter; printf("counter = %d\n",counter); printf("Class average is %.2f\n", average); } else printf("No grades were entered\n");

return(0); return(0);

} }

Output:

Chapter 5 – Appendix

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5) Write nests of loops that cause the following output to be displayed: Inner loop 123456

Outer loop

1 2 3 4 5 6

* ** *** **** ***** ******

Solution: Using for (nested loop)

Using for and while (nested loop)

Using while (nested loop)

Using do-while (nested loop)

#include<stdio.h>

#include<stdio.h>

#include<stdio.h>

#include<stdio.h>

int main(void) { int i; int j;

int main(void) { int i; int j;

int main(void) { int i; int j;

int main(void) { int i; int j;

for(i=1; i<=6; i++) { for(j=1; j<=i; j++) printf("*"); printf("\n"); }

for(i=1; i<=6; i++) { j=1; while(j<=i) { printf("*"); j++; } printf("\n"); }

return(0); }

i=1; do { j=1; do { printf("*"); j++; } while(j<=i); printf("\n"); i++; } while(i<=6);

i=1; while(i<=6) { j=1; while(j<=i) { printf("*"); j++; } printf("\n"); i++; }

return(0); }

return(0);

return(0); }

Output:

}

Chapter 5 – Appendix

5/18/2008

6) Write nests of loops that cause the following output to be displayed: 0 0 0 0 0 0 0 0 0

1 1 1 1 1 1 1

2 2 3 2 3 4 2 3 2

Solution: #include<stdio.h> int main(void) { int outer,inner,k,m=0; for(outer=0;outer<=8;outer++) { if(outer<=4) k=outer; else { m+=2; k=outer-m; } for(inner=0;inner<=k;inner++) printf("%2d",inner); printf("\n"); } return(0); }

Output:

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