C Language : Call by Reference Vs Call by Value [已解决]

蜡笔小烦

1. /*
   
2. Homework for Chapter 5:
   
3. Write a C program that accept a student quiz mark qm (per qt),
   
4. before calculating the actual contribution (in %) of the quiz, qcp,to his or her total mark (per 100).
   
5. e.g.    Ali's obtained 45/60 (qm = 45, qt = 60) for his quiz.
   
6.   Since the quiz contributes to 15% (qcp= 15) towards the total mark,
   
7.   his current total mark is:??    qm*(qcp/qt) = 45 x (15/60) =  11.25
   
8. The above formula must be written as a separate C function.
   
9. Prepare two version of the program: use "call by value" for the 1st version, and "call by reference" for the 2nd version.
   
10. */
    
11. #include<stdio.h>
    
12. float calculateMark(float fullQuiz,float quizMark,float conMark)
    
13. {
    
14. float totalMark=(quizMark/fullQuiz)*conMark;
    
15. return totalMark;
    
16. }
    
17. ////////////////////////////////////////////////
    
18. //////////1st Version: Call By Value////////////
    
19. ////////////////////////////////////////////////
    
20. void main()
    
21. {
    
22. float qm,qt,qcp,totalMark;
    
23. char percent[]="%";
    
24. printf("lease enter the full quiz mark: ";
    
25. scanf("%f",&qt);
    
26. printf("lease enter your quiz mark: ";
    
27. scanf("%f",&qm);
    
28. printf("The quiz contributes to how many mark towards the total mark? ";
    
29. scanf("%f",&qcp);
    
30. totalMark=calculateMark(qt,qm,qcp);
    
31. printf("Your actual contribution (in %s) of the quiz to your total mark is %.2f%s.\n",percent,totalMark,percent);
    
32. }

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1. /*
   
2. Homework for Chapter 5:
   
3. Write a C program that accept a student quiz mark qm (per qt),
   
4. before calculating the actual contribution (in %) of the quiz, qcp,to his or her total mark (per 100).
   
5. e.g.    Ali's obtained 45/60 (qm = 45, qt = 60) for his quiz.
   
6.   Since the quiz contributes to 15% (qcp= 15) towards the total mark,
   
7.   his current total mark is:??    qm*(qcp/qt) = 45 x (15/60) =  11.25
   
8. The above formula must be written as a separate C function.
   
9. Prepare two version of the program: use "call by value" for the 1st version, and "call by reference" for the 2nd version.
   
10. */
    
11. #include<stdio.h>
    
12. float calculateMark(float *pointer_qt,float *pointer_qm,float *pointer_qcp)
    
13. {
    
14. float quizMark,fullQuiz,conMark,totalMark;
    
15. printf("lease enter the full quiz mark: ";
    
16. scanf("%f",&fullQuiz);
    
17. printf("lease enter your quiz mark: ";
    
18. scanf("%f",&quizMark);
    
19. printf("The quiz contributes to how many mark towards the total mark? ";
    
20. scanf("%f",&conMark);
    
21. *pointer_qt=fullQuiz;
    
22. *pointer_qm=quizMark;
    
23. *pointer_qcp=conMark;
    
24. totalMark=(*pointer_qm / *pointer_qt) * *pointer_qcp;
    
25. return totalMark;
    
26. }
    
27. ////////////////////////////////////////////////
    
28. //////////2nd Version: Call By Reference////////
    
29. ////////////////////////////////////////////////
    
30. void main()
    
31. {
    
32. float qm,qt,qcp,totalMark;
    
33. char percent[]="%";
    
34. totalMark=calculateMark(&qt,&qm,&qcp);
    
35. printf("Your actual contribution (in %s) of the quiz to your total mark is %.2f%s.\n",percent,totalMark,percent);
    
36. }

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我不确定Call by reference的version是不是这样做，
总觉得怪怪的。
请大家指点迷津。。。
谢谢！~

[ 本帖最后由 蜡笔小烦 于 18-10-2008 03:29 PM 编辑 ]

cheng1986

給我的話我會這樣solve...
把fullQuiz, quizMark, conMark的address傳過去
其實很沒有意義...

把totalMark的address傳過去的話
不用return也可以改掉totalMark的value...

1. #include<stdio.h>
   
2. float calculateMark(float fullQuiz, float quizMark, float conMark, float *totalMark)
   
3. {
   
4. *totalMark=(quizMark / fullQuiz) * conMark;
   
5. }
   
6. ////////////////////////////////////////////////
   
7. //////////2nd Version: Call By Reference////////
   
8. ////////////////////////////////////////////////
   
9. void main()
   
10. {
    
11. float quizMark, fullQuiz, conMark, totalMark;
    
12. char percent[]="%";
    
13. printf("lease enter the full quiz mark: ";
    
14. scanf("%f",&fullQuiz);
    
15. printf("lease enter your quiz mark: ";
    
16. scanf("%f",&quizMark);
    
17. printf("The quiz contributes to how many mark towards the total mark? ";
    
18. scanf("%f",&conMark);
    
19. calculateMark(fullQuiz, quizMark, conMark, &totalMark);
    
20. printf("Your actual contribution (in %s) of the quiz to your total mark is %.2f%s.\n",percent,totalMark,percent);
    
21. }

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[ 本帖最后由 cheng1986 于 6-10-2008 01:31 PM 编辑 ]

蜡笔小烦

原帖由 cheng1986 于 6-10-2008 01:30 PM 发表
給我的話我會這樣solve...
把fullQuiz, quizMark, conMark的address傳過去
其實很沒有意義...

把totalMark的address傳過去的話
不用return也可以改掉totalMark的value...

#include
float calculateMark(fl ...

对~
我就是觉得我那做法很多余~
还学不精的说
还请多多指教！~

谢谢你~

ps:有个warning说 'calculateMark' : must return a value
不用管它吗？

cheng1986

function type要改成void.....

蜡笔小烦

原帖由 cheng1986 于 7-10-2008 01:31 AM 发表
function type要改成void.....

对啊~
没有察觉到
不好意思
谢谢！~