成绩计划问题 [英] Grade Program Question

问题描述

我正在做一个书中练习的程序。关键在于创建一个程序，该程序将从用户那里获取数字等级并将其转换为

字母等级（是的，非常简单）。我尝试将while循环合并到

让程序继续运行，这样用户可以继续输入成绩

来获取字母数字，但是我收效甚微，只是保持不变创造

无限循环。我的问题是：


我怎么能修改我的程序来读取输入的最后一个字母并添加+

或 - 到字母等级的末尾..

1-3 = -

4-6 =< blank>

7-9 = +


到目前为止，这是我的程序：


#include< stdio.h>

#include< string.h> < br $>
main（）

{

int letter_grade;

int numeric_grade;

int等级[100];


（无效）printf（"输入数字等级："）;

（无效）fgets（等级，sizeof（等级），stdin）;

（无效）sscanf（等级，％d，& numeric_grade）;


if（numeric_grade< = 60）

letter_grade =" F"

if（numeric_grade> 60&& numeric_grade< = 70）

letter_grade =" D";

if（numeric_grade> 70&& numeric_grade< = 80）

letter_grade =" C";

if（numeric_g rade> 80&& numeric_grade< = 90）

letter_grade =" B";

if（numeric_grade> 90&& numeric_grade< = 100）

letter_grade =" A";


（无效）printf（信用等级为：％s \ n，letter_grade）;

}

---

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解决方案

" Henry" <他***** @ knology.net>在消息中写道

news：vl ************ @ corp.supernews.com ...

我正在做这个程序在一本书中锻炼。关键在于
创建一个程序，该程序将从用户获取数字等级并将其转换为
字母等级（是的非常简单）。我试图结合while循环来保持程序连续运行，这样用户可以继续输入
等级来获得字母编号，但是我没有成功，只是不断创建
无限循环。我的问题是：

如何修改我的程序来读取输入的最后一个字母，并在字母等级的末尾添加
+或 -
1- 3 = -
4-6 =< blank>
7-9 = +

这是我的程序到目前为止：

#include< ; stdio.h>
#include< string.h>
main（）


它应该是 int main（）"。如果C书写main（），那就错了，

它不符合ANSI标准。


{
int letter_grade;
int numeric_grade;
int grade [100];

（void）printf（" Enter the Numeric grade："）;
（void ）fgets（grade，sizeof（grade），stdin）;
（void）sscanf（grade，"％d"，& numeric_grade）;


（无效）是不必要的，只是令人困惑。你可以写


printf（输入数字等级：）;

得到（等级，sizeof（等级），stdin）;

sscanf（grade，"％d"，& numeric_grade）;

if（numeric_grade< = 60）
letter_grade =" F" ;;


F是字符串文字，你的letter_grade是是一个整数。你只能用
写


letter_grade =''F'';

if（numeric_grade> 60&& numeric_grade< = 70）
letter_grade =" D" ;;
if（numeric_grade> 70&& numeric_grade< = 80）
letter_grade =" C" ;;
if（numeric_grade> 80&& numeric_grade< = 90）
letter_grade =" B" ;;
if（numeric_grade>&& numeric_grade< = 100）
letter_grade =" A" ;;

（void）printf（" The Letter grade is ：％s \ n"，letter_grade）;
}


你不能将letter_grade用于％s。请记住，letter_grade不是字符串。

你可以写


printf（字母等级是：％c \ n，letter_grade）;

-

Jeff

Jeff写道：< blockquote class =post_quotes>" Henry" <他***** @ knology.net>在消息中写道
新闻：vl ************ @ corp.supernews.com ...

我正在做这个程序在一本书中锻炼。关键是要创建

一个程序，它将取一个用户的数字等级并将其转换为

a < br>

字母等级（是的，非常容易）。我尝试将while循环结合起来以保持程序不断运行，这样用户可以继续输入

来获取字母编号，但我没有成功，只是不断创建无限循环。我的问题是：

如何修改程序以读取输入的最后一个字母并添加

或 - 到字母等级的末尾..
1-3 = -
4-6 =< blank>
7-9 = +

#include< stdio.h>
#include< string.h>
main（）

它应该是 int main（）"。如果C书写main（），那就错了，
它不符合ANSI标准。

{
int letter_grade;
int numeric_grade;
int grade [100];

（void）printf（" Enter the Numeric grade："）;
（void）fgets（等级） ，sizeof（grade），stdin）;
（void）sscanf（grade，"％d"，& numeric_grade）;

（void）是不必要的，这简直令人困惑。你可以写

printf（输入数字等级：）;
得到（等级，sizeof（等级），stdin）;

显然你的意思是：fgets（...）;

sscanf（等级，％d，& numeric_grade）;

if（numeric_grade< = 60）
letter_grade =" F";

" F"是字符串文字，你的letter_grade是是一个整数。你只能写

letter_grade =''F'';

if（numeric_grade> 60&& numeric_grade< = 70）
letter_grade =" D";

if（numeric_grade> 70&& numeric_grade< = 80）
letter_grade =" C";

if（numeric_grade> 80&& numeric_grade< = 90）
letter_grade =" B";

if（numeric_grade> 90&& numeric_grade< = 100） letter_grade =" A" ;;

（void）printf（字母等级为：％s \ n，letter_grade）;
}

你不能将letter_grade用于％s。请记住，letter_grade不是字符串。
你可以写

printf（字母等级是：％c \ nn，letter_grade）;

HTH，

--ag


ps printf上的void命令，等等。人。有时需要使用
过于激进的类似lint的工具停止抱怨因为

丢弃的结果值 - 并且，如OR状态，没有其他

理由。


-

Artie Gold - 德克萨斯州奥斯汀

>它应该是 int main（）"。如果C书写main（），那就错了，

它不符合ANSI标准。


即使在K& R第二版中它还没有提及关于int的任何内容

main（）

（无效） ）是不必要的，这只是令人困惑。你可以写

printf（输入数字等级：）;
得到（等级，sizeof（等级），stdin）;
sscanf（等级，" ％d"，& numeric_grade）;


我只是在做什么实用C告诉我......


" F"是字符串文字，你的letter_grade是是一个整数。你只能写


letter_grade =''F'';


我尝试了两种方式，两种方式都有效。

你不能将letter_grade用于％s。请记住，letter_grade不是字符串。
你可以写

printf（字母等级是：％c \ nn，letter_grade）;

不是由字符组成的字符串吗？从技术上讲，两种方式都是正确的吗？

.....

---

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I was doing this program for an exercise in a book. The point was to create
a program that would take a numerical grade from a user and convert it to a
letter grade (yeah really easy). I tried to incorporate the while loop to
keep the program running continuously so the user could keep entering grades
to get the letter number, but I had little success and just kept creating
infinite loops. My question is:

How could I modify my program to read the last letter of the input and add +
or - to the end of the letter grade..
1-3 = -
4-6 = <blank>
7-9 = +

Here is my program so far:

#include <stdio.h>
#include <string.h>
main()
{
int letter_grade;
int numeric_grade;
int grade[100];

(void)printf("Enter the Numeric grade: ");
(void)fgets(grade, sizeof(grade), stdin);
(void)sscanf(grade, "%d", &numeric_grade);

if (numeric_grade <= 60)
letter_grade = "F";
if (numeric_grade > 60 && numeric_grade <= 70)
letter_grade = "D";
if (numeric_grade > 70 && numeric_grade <= 80)
letter_grade = "C";
if (numeric_grade > 80 && numeric_grade <= 90)
letter_grade = "B";
if (numeric_grade > 90 && numeric_grade <= 100)
letter_grade = "A";

(void)printf("The Letter grade is: %s\n", letter_grade);
}
---
Outgoing mail is certified Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
Version: 6.0.515 / Virus Database: 313 - Release Date: 9/1/2003

解决方案

"Henry" <he*****@knology.net> wrote in message
news:vl************@corp.supernews.com...

I was doing this program for an exercise in a book. The point was to create a program that would take a numerical grade from a user and convert it to a letter grade (yeah really easy). I tried to incorporate the while loop to
keep the program running continuously so the user could keep entering grades to get the letter number, but I had little success and just kept creating
infinite loops. My question is:

How could I modify my program to read the last letter of the input and add + or - to the end of the letter grade..
1-3 = -
4-6 = <blank>
7-9 = +

Here is my program so far:

#include <stdio.h>
#include <string.h>
main()
It should be " int main() ". If you C book write "main()", it is wrong and
it does not follow the ANSI standard.

{
int letter_grade;
int numeric_grade;
int grade[100];

(void)printf("Enter the Numeric grade: ");
(void)fgets(grade, sizeof(grade), stdin);
(void)sscanf(grade, "%d", &numeric_grade);
The (void) is unnecessary, it is just confusing. You can write

printf("Enter the Numeric grade: ");
gets(grade, sizeof(grade), stdin);
sscanf(grade, "%d", &numeric_grade);

if (numeric_grade <= 60)
letter_grade = "F";
"F" is string literal, and your "letter_grade" is an integer. You can only
write

letter_grade = ''F'';
if (numeric_grade > 60 && numeric_grade <= 70)
letter_grade = "D"; if (numeric_grade > 70 && numeric_grade <= 80)
letter_grade = "C"; if (numeric_grade > 80 && numeric_grade <= 90)
letter_grade = "B"; if (numeric_grade > 90 && numeric_grade <= 100)
letter_grade = "A";

(void)printf("The Letter grade is: %s\n", letter_grade);
}
You can''t use letter_grade for %s. Remember, letter_grade IS NOT a string.
You can write

printf("the letter grade is: %c\n", letter_grade);

--
Jeff

Jeff wrote:

"Henry" <he*****@knology.net> wrote in message
news:vl************@corp.supernews.com...

I was doing this program for an exercise in a book. The point was to
create

a program that would take a numerical grade from a user and convert it to

a

letter grade (yeah really easy). I tried to incorporate the while loop to
keep the program running continuously so the user could keep entering

grades

to get the letter number, but I had little success and just kept creating
infinite loops. My question is:

How could I modify my program to read the last letter of the input and add

+

or - to the end of the letter grade..
1-3 = -
4-6 = <blank>
7-9 = +

Here is my program so far:

#include <stdio.h>
#include <string.h>
main()

It should be " int main() ". If you C book write "main()", it is wrong and
it does not follow the ANSI standard.

{
int letter_grade;
int numeric_grade;
int grade[100];

(void)printf("Enter the Numeric grade: ");
(void)fgets(grade, sizeof(grade), stdin);
(void)sscanf(grade, "%d", &numeric_grade);

The (void) is unnecessary, it is just confusing. You can write

printf("Enter the Numeric grade: ");
gets(grade, sizeof(grade), stdin);

Obviously you meant: fgets(...);
sscanf(grade, "%d", &numeric_grade);

if (numeric_grade <= 60)
letter_grade = "F";

"F" is string literal, and your "letter_grade" is an integer. You can only
write

letter_grade = ''F'';

if (numeric_grade > 60 && numeric_grade <= 70)
letter_grade = "D";

if (numeric_grade > 70 && numeric_grade <= 80)
letter_grade = "C";

if (numeric_grade > 80 && numeric_grade <= 90)
letter_grade = "B";

if (numeric_grade > 90 && numeric_grade <= 100)
letter_grade = "A";

(void)printf("The Letter grade is: %s\n", letter_grade);
}

You can''t use letter_grade for %s. Remember, letter_grade IS NOT a string.
You can write

printf("the letter grade is: %c\n", letter_grade);

HTH,
--ag

p.s. The cast to `void'' on printf, et. al. is sometimes necessary to
make overly aggressive lint-like tools stop complaining because of
the discarded result value -- and, as the OR states, for no other
reason.

--
Artie Gold -- Austin, Texas

> It should be " int main() ". If you C book write "main()", it is wrong and

it does not follow the ANSI standard.
Well even in K&R second edition it has yet to mention anything about int
main()
The (void) is unnecessary, it is just confusing. You can write

printf("Enter the Numeric grade: ");
gets(grade, sizeof(grade), stdin);
sscanf(grade, "%d", &numeric_grade);
I''m just doing what "Practical C" is telling me to..

"F" is string literal, and your "letter_grade" is an integer. You can only write

letter_grade = ''F'';

I tried it both ways , and both ways worked.

You can''t use letter_grade for %s. Remember, letter_grade IS NOT a string.
You can write

printf("the letter grade is: %c\n", letter_grade);

Isn''t a string made up of chars? So technically both ways would be right?
.....
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