我是一名新的C编程学习者，我正在尝试制作一个程序，该程序将根据成绩输入和科目学分计算学生的GPA。
我遇到的问题是我想将输入的主题数限制为仅2个到6个。
另一个问题是我想限制用户只能输入1到100之间的整数，而不是其他任何关键字，特殊字符（EOF）
我已将“ ###”放在注释行中，我需要进行这些修改。
#include <stdio.h>
int main(void) {
// input user input -- hopefully a number
// temp used to collect garbage characters
// status did the user enter a number?
// counter for keeping track of loop repetition
// no no. of subjects to be entered by user.
// credits credits per subject
// grades grades acheived in each subject (1 to 100).
// grade_value for holding the value of each subject grade (for ex; 80 to hundred is 4.0)
// grade_points Grade points for each subject (credits * grade_value)
// sum sum of total grade points
int counter = 1, subjects, no, credits, grades, status, temp;
float grade_value, grade_point, sum;
printf("Enter number of subjects you took for current semester: ");
status = scanf("%d", & no);
// ### I want to limit this integer input to be >=2 && <=6.
while (status != 1) {
while ((temp = getchar()) != EOF && temp != '\n');
if ((temp < 2) && (temp > 6));
break;
printf("Invalid input... please enter the number of subject again: ");
status = scanf("%d", & no);
}
// ### I want to be this input to block other character inputs than integer from 1 to hundred.
while (counter <= no) {
printf("\nEnter Subject %d grades separated with credits \n", counter);
scanf("%d %d", & grades, & credits);
if ((grades > 0) && (grades <= 29)) {
grade_value = 0;
printf("Grade value for subject %d is: %.2f", counter, grade_value);
grade_point = credits * grade_value;
printf("\nGrade point for subject %d is: %.2f", counter, grade_point);
sum = sum + grade_point;
++counter;
} else if ((grades >= 30) && (grades <= 34)) {
grade_value = 0.67;
printf("Grade value for subject %d is: %.2f", counter, grade_value);
grade_point = credits * grade_value;
printf("\nGrade point for subject %d is: %.2f", counter, grade_point);
sum = sum + grade_point;
++counter;
} else if ((grades >= 35) && (grades <= 39)) {
grade_value = 1;
printf("Grade value for subject %d is: %.2f", counter, grade_value);
grade_point = credits * grade_value;
printf("\nGrade point for subject %d is: %.2f", counter, grade_point);
sum = sum + grade_point;
++counter;
} else if ((grades >= 40) && (grades <= 44)) {
grade_value = 1.33;
printf("Grade value for subject %d is: %.2f", counter, grade_value);
grade_point = credits * grade_value;
printf("\nGrade point for subject %d is: %.2f", counter, grade_point);
sum = sum + grade_point;
++counter;
} else if ((grades >= 45) && (grades <= 49)) {
grade_value = 1.67;
printf("Grade value for subject %d is: %.2f", counter, grade_value);
grade_point = credits * grade_value;
printf(" \nGrade point for subject %d is: %.2f", counter, grade_point);
sum = sum + grade_point;
++counter;
} else if ((grades >= 50) && (grades <= 54)) {
grade_value = 2;
printf("Grade value for subject %d is: %.2f", counter, grade_value);
grade_point = credits * grade_value;
printf("\nGrade point for subject %d is: %.2f", counter, grade_point);
sum = sum + grade_point;
++counter;
} else if ((grades >= 55) && (grades <= 59)) {
grade_value = 2.33;
printf("Grade value for subject %d is: %.2f", counter, grade_value);
grade_point = credits * grade_value;
printf("\nGrade point for subject %d is: %.2f", counter, grade_point);
sum = sum + grade_point;
++counter;
} else if ((grades >= 60) && (grades <= 64)) {
grade_value = 2.67;
printf("Grade value for subject %d is: %.2f", counter, grade_value);
grade_point = credits * grade_value;
printf("\nGrade point for subject %d is: %.2f", counter, grade_point);
sum = sum + grade_point;
++counter;
} else if ((grades >= 65) && (grades <= 69)) {
grade_value = 3;
printf("Grade value for subject %d is: %.2f", counter, grade_value);
grade_point = credits * grade_value;
printf("\nGrade point for subject %d is: %.2f", counter, grade_point);
sum = sum + grade_point;
++counter;
} else if ((grades >= 70) && (grades <= 74)) {
grade_value = 3.33;
printf("Grade value for subject %d is: %.2f", counter, grade_value);
grade_point = credits * grade_value;
printf("\nGrade point for subject %d is: %.2f", counter, grade_point);
sum = sum + grade_point;
++counter;
} else if ((grades >= 75) && (grades <= 79)) {
grade_value = 3.67;
printf("Grade value for subject %d is: %.2f", counter, grade_value);
grade_point = credits * grade_value;
printf("\nGrade point for subject %d is: %.2f", counter, grade_point);
sum = sum + grade_point;
++counter;
} else if ((grades >= 80) && (grades <= 100)) {
grade_value = 4;
printf("Grade value for subject %d is: %.2f", counter, grade_value);
grade_point = credits * grade_value;
printf("\nGrade point for subject %d is: %.2f", counter, grade_point);
sum = sum + grade_point;
++counter;
}
// To print a message if user doesnt enter an integer varying from 1 to 100.
else {
printf("\n Error Grade input, Please Key in Again. (1 to 100 only.)");
}
}
printf("\n");
printf("\n");
printf("\nThe GPA is: %.2f", sum);
if (sum <= 49) {
printf("\nYou can register for 2 subjects for next semester.");
} else if ((sum >= 50) && (sum >= 79)) {
printf("\nYou can register for 5 subjects for next semester.");
} else if ((sum >= 80) && (sum <= 100)) {
printf("\nYou can register for 6 subjects for next semester.");
}
printf("\n");
printf("\n");
printf("\n_______________________________________________________");
printf("\nEnd of program");
return 0;
}
答案 0 :(得分：0)
do-while循环非常适合需要至少调用一次且可以消除重复代码的循环。读取可能来自键盘的非结构化输入（可能会调用./a.out < text.txt）实际上是很棘手的事情。
幸运的是，C常见问题解答有很多建议，例如http://c-faq.com/stdio/scanfprobs.html。但是，如果没有功能，它将变得非常艰巨。通过有限的测试，我很确定这是读取第一个变量的可靠方法。
#include <stdio.h> /* fgets sscanf */
#include <stdlib.h> /* EXIT_ printf fprintf */
#include <string.h> /* strlen */
int main(void) {
int no;
/* Input number, no \in [2, 6], and make sure that the read cursor is on
the next line. */
do {
char buffer[80];
size_t len;
printf("Enter number of subjects you took for current semester, [2, 6]: ");
/* On the advice of http://c-faq.com/stdio/scanfprobs.html, this reads
into a buffer first. */
if(!fgets(buffer, sizeof buffer, stdin)) {
if(feof(stdin)) {
fprintf(stderr, "Premature EOF.\n");
} else {
/* On IEEE Std 1003.1-2001-conforming systems, this will print
a meaningful error. */
perror("stdin");
}
/* Can't really do anything interactive once stdin has a read
error. */
return EXIT_FAILURE;
}
/* This is always going to be true, but segfaults if not. Paranoid. */
if(!(len = strlen(buffer))) continue;
/* Normally fgets stores a '\n' at the end; check. */
if(buffer[len - 1] != '\n') {
/* Check if the length of the buffer is big enough to hold input. */
if(len >= sizeof buffer - 1) {
int c;
fprintf(stderr, "Line too long.\n");
/* Flush whole line. http://c-faq.com/stdio/stdinflush2.html */
while((c = getchar()) != '\n') {
if(c != EOF) continue;
if(feof(stdin)) fprintf(stderr, "Premature EOF.\n");
else perror("stdin");
return EXIT_FAILURE;
}
continue;
} else {
/* Non-POSIX lines, eg, file without '\n' terminating. */
fprintf(stderr, "Note: line without line break detected.\n");
}
}
/* Parse buffer for a number that's between [2, 6]. Ignore the rest. */
if(sscanf(buffer, "%d", &no) != 1 || no < 2 || no > 6) {
fprintf(stderr, "Invalid input.\n");
continue;
}
/* Now no \in [2, 6]. */
printf("no: %d\n", no);
break;
} while(1);
return EXIT_SUCCESS;
}
如果您的老师仅输入格式正确的数字，则可能可以摆脱其中的一部分。
答案 1 :(得分：0)
您的代码有一些语义错误。切勿在{{1}}语句后加上分号。您也可以通过间隔代码块来编写更清晰的代码。更好的是，您可以重构一些块并将其封装到函数中。除此之外，请参阅以下解决方案：
对于第一个问题，只需几行代码即可解决。好的做法是，用户在输入输入之前先了解输入的限制：
if另一个问题可以通过使用Standard C库来解决。只是/* ### I want to limit this integer input to be >=2 && <=6. */
#include <stdio.h>
int main()
{
int no;
do {
printf("Enter number of subjects you took for current semester (2~6): ");
scanf("%d", &no);
} while (no < 2 || no > 6);
return 0;
}
和#include <string.h>
#include <ctype.h>