i need help with my work, here is what my professor said.good
i need help with my work, here is what my professor said.<br/>good
start – but check with a tutor or try our tutoring area in class for a little more help in understanding ….
Lab Execution: … need to show all inputs in STDIN and run the code with several test and show all output.
Modifies the code: … this was not done
Describes char array: … The array runs from position 0 to 99 by default which is 100 positions. Also mention that the null character is always at the end of a char array and so there is one less storage space available.
The rubric and my work is attached
Lab 62. Modify the code to be able to input an undetermined number of students. You willstill only have 3 exams for each student. Support your experimentation with screencaptures of executing the new code.//C code//This program will calculate the average of 3 exams for 5 students.//Developer: Faculty CMIS102//Date: Jan 31, XXXX#include <stdio.h>int main (){/* variable definition: */char StudentName[100];float ExamValue, Sum, Avg;int students,exams,choice;// Loop through 5 Studentswhile(1){//reset Sum to 0Sum =0.0;printf(“Enter Student Name: “);scanf(“%s”, StudentName);//Nested Loop for Examsfor (exams=0; exams < 3; exams++){printf (“Enter exam grade:“);scanf(“%f”, &ExamValue);Sum += ExamValue;}Avg = Sum/3.0;printf( “Average for %s is %f\n”,StudentName,Avg);printf(“Enter 1 to continue or 0 to exit: “);scanf(“%d”,&choice);if(choice==0)break;}printf(“Bye!!!”);return 0;
}3. Prepare a new test table with at least 3 distinct test cases listing input and expectedoutput for the code you created after step 1.Test CaseInputExpected Output1Student Name=JonExamValue1=50ExamValue2=50ExamValue3=50Choice=0Average for Jon is50.000000Bye!!!2Student Name=JonExamValue1=50ExamValue2=50ExamValue3=50Choice=1Student Name=SmithExamValue1=33ExamValue2=45ExamValue3=65Choice=1Student Name=MarkExamValue1=30ExamValue2=50Average for Jon is50.000000Average for Smithis 47.666668Average for Mark is40.000000Bye!!!
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