C语言 如何在 C 中使用带有用户输入的循环函数？
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How to use a loop function with user input in C?
提问by Markus Andrew White
Here is the average program I created.
这是我创建的平均程序。
#include <stdio.h>
#include <stdlib.h>
int main()
{
float f1,f2,f3;
/* Program to calculate averages. */
/*Asks for the numbers.*/
printf(" Please enter three numbers.\n");
printf ("\t" "First number please.\n");
scanf("%f", &f1);
printf ("\t" "Second number please.\n");
scanf ("%f", &f2);
printf("\t" "Third number please.\n");
scanf("%f", &f3);
/* Now it averages it.*/
printf(" Thank you, wait one.\n");
printf(" Excellent, your sum is.\n");
printf("%f""\n", f1+f2+f3);
printf("Your average of the sum is now!!!!\n");
printf("%f", (f1+f2+f3)/3);
return 0;
}
Now would I turn this into a do-while? Or an if else?
现在我会把它变成一个 do-while 吗？还是如果？
回答by Jonathan Leffler
If you want to repeat the whole entry and averaging process, you can wrap a loop around the code:
如果要重复整个输入和平均过程，可以在代码周围环绕一个循环：
#include <stdio.h>
int main(void)
{
float f1,f2,f3;
while (1)
{
printf("Please enter three numbers.\n");
printf("\tFirst number please.\n");
if (scanf("%f", &f1) != 1)
break;
printf("\tSecond number please.\n");
if (scanf("%f", &f2) != 1)
break;
printf("\tThird number please.\n");
if (scanf("%f", &f3) != 1)
break;
printf("Your sum is %f\n", f1+f2+f3);
printf("Your average is %f\n", (f1+f2+f3)/3);
}
return 0;
}
Note that this code checks the return value from scanf()each time it is used, breaking the loop if there's a problem. There's no need for string concatenation, and a single printf()can certainly print a string and a value.
请注意，此代码在scanf()每次使用时都会检查返回值，如果出现问题则中断循环。不需要字符串连接，单个printf()肯定可以打印一个字符串和一个值。
That's a simple first stage; there are more elaborate techniques that could be used. For example, you could create a function to prompt for and read the number:
这是一个简单的第一阶段；可以使用更复杂的技术。例如，您可以创建一个函数来提示和读取数字：
#include <stdio.h>
static int prompt_and_read(const char *prompt, float *value)
{
printf("%s", prompt);
if (scanf("%f", value) != 1)
return -1;
return 0;
}
int main(void)
{
float f1,f2,f3;
while (printf("Please enter three numbers.\n") > 0 &&
prompt_and_read("\tFirst number please.\n", &f1) == 0 &&
prompt_and_read("\tSecond number please.\n", &f2) == 0 &&
prompt_and_read("\tThird number please.\n", &f3) == 0)
{
printf("Your sum is %f\n", f1+f2+f3);
printf("Your average is %f\n", (f1+f2+f3)/3);
}
return 0;
}
If you want to get away from a fixed set of three values, then you can iterate until you encounter EOF or an error:
如果你想摆脱一组固定的三个值，那么你可以迭代，直到遇到 EOF 或错误：
#include <stdio.h>
static int prompt_and_read(const char *prompt, float *value)
{
printf("%s", prompt);
if (scanf("%f", value) != 1)
return -1;
return 0;
}
int main(void)
{
float value;
float sum = 0.0;
int num = 0;
printf("Please enter numbers.\n");
while (prompt_and_read("\tNext number please.\n", &value) == 0)
{
sum += value;
num++;
}
if (num > 0)
{
printf("You entered %d numbers\n", num);
printf("Your sum is %f\n", sum);
printf("Your average is %f\n", sum / num);
}
return 0;
}
You might also decide to replace the newline at the ends of the prompt strings with a space so that the value is typed on the same line as the prompt.
您可能还决定用空格替换提示字符串末尾的换行符，以便在提示所在的同一行中键入该值。
If you want to check whether to repeat the calculation, you can use a minor variant on the first or second versions of the code:
如果要检查是否重复计算，可以在代码的第一个或第二个版本上使用次要变体：
#include <stdio.h>
static int prompt_and_read(const char *prompt, float *value)
{
printf("%s", prompt);
if (scanf("%f", value) != 1)
return -1;
return 0;
}
static int prompt_continue(const char *prompt)
{
printf("%s", prompt);
char answer[2];
if (scanf("%1s", answer) != 1)
return 0;
if (answer[0] == 'y' || answer[0] == 'Y')
{
int c;
while ((c = getchar()) != EOF && c != '\n') // Gobble to newline
;
return 1;
}
return 0;
}
int main(void)
{
float f1,f2,f3;
while (printf("Please enter three numbers.\n") > 0 &&
prompt_and_read("\tFirst number please.\n", &f1) == 0 &&
prompt_and_read("\tSecond number please.\n", &f2) == 0 &&
prompt_and_read("\tThird number please.\n", &f3) == 0)
{
printf("Your sum is %f\n", f1+f2+f3);
printf("Your average is %f\n", (f1+f2+f3)/3);
if (prompt_continue("Do you want to try again?") == 0)
break;
}
return 0;
}
回答by Houssem Badri
You can do this:
你可以这样做：
int main()
{
float number, sum=0.0f;
int index=0;
do
{
printf ("\t" "Enter number please.\n"); //Asking for a number from user
scanf("%f", &number); //Getting a number from a user
sum+=number; //Add number entered to the sum
i++;
} while (i < 3);
printf("Excellent, your average is %f\n", sum/3);
return 0;
}
回答by Civa
#include <stdio.h>
#include <stdlib.h>
int main()
{
float f1,f2,f3;
char c='Y';
/* Program to calculate averages. */
/*Asks for the numbers.*/
do
{
printf(" Please enter three numbers.\n");
printf ("\t" "First number please.\n");
scanf("%f", &f1);
printf ("\t" "Second number please.\n");
scanf ("%f", &f2);
printf("\t" "Third number please.\n");
scanf("%f", &f3);
/* Now it averages it.*/
printf(" Thank you, wait one.\n");
printf(" Excellent, your sum is.\n");
printf("%f""\n", f1+f2+f3);
printf("Your average of the sum is now!!!!\n");
printf("%f", (f1+f2+f3)/3);
printf ("Do you wana continue [Y/N]...\n");
scanf("%c", &c);
}while(c!='N'&&c!='n');
return 0;
}
