问题描述
这是我制作的程序:
#include <stdio.h>
int main(void)
{
//Put variables here
int space = 0;
int num_of_rows = 0;
int p = 1;
int t = 0;
char alphabet[100] = {'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'};
get_and_validate_user_input(num_of_rows);
printf(" The number of rows are : %d",num_of_rows);
}
void get_and_validate_user_input(int num_of_rows)
{
//Input validation
while (1)
{
printf("Enter the number of rows : ");
//Getting the input from the user
scanf("%d",&num_of_rows);
if (num_of_rows <= 0)
{
printf("Negative numbers are not allowed here \n");
}
else
{
break;
}
}
}
预期输入: 输入行数:5
预期的输出: 行数是:5
现在我的问题是:我的程序做错了什么?我该如何解决这个问题?
解决方法
您的问题是 get_and_validate_user_input 修改了 local 变量 num_of_rows =>值保持为 main 0 >
如果您还希望从 get_and_validate_user_input 中修改 num_of_rows ,请按地址而不是按值提供它,例如(未使用的变量已注释,请修改 get_and_validate_user_input中的检查与消息兼容):
#include <stdio.h>
#include <stdlib.h>
void get_and_validate_user_input(int * num_of_rows);
int
main(void){
//Put variables here
//int space=0;
int num_of_rows=0;
//int p=1;
//int t=0;
//char alphabet[100]={'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'};
get_and_validate_user_input(&num_of_rows);
printf(" The number of rows is : %d\n",num_of_rows);
return 0;
}
void get_and_validate_user_input(int * num_of_rows){
//Input validation
while(1){
printf("Enter the number of rows : ");
//Getting the input from the user
if (scanf("%d",num_of_rows) != 1) {
int c;
puts("Invalid input");
// invalid input,bypass all the line
while ((c = getchar()) != '\n') {
if (c == EOF) {
puts("EOF,abort");
exit(-1);
}
}
}
else if(*num_of_rows < 0){ // 0 seems allowed,else update message
printf("Negative numbers are not allowed here \n");
}
else {
break;
}
}
}
编译和执行:
/tmp % gcc -Wall c.c
/tmp % ./a.out
Enter the number of rows : 12
The number of rows is : 12
/tmp % ./a.out
Enter the number of rows : aze
Invalid input
Enter the number of rows : -2
Negative numbers are not allowed here
Enter the number of rows : 12
The number of rows is : 12
/tmp %
,
编译器看不到调用点
get_and_validate_user_input(num_of_rows);
如何声明函数get_and_validate_user_input
。因此,编译器发出一条消息。它期望该函数默认情况下具有返回类型int
,但是当遇到函数定义时,它将看到返回类型为void
。
将函数声明放置在main之前。
函数中未使用传递的参数的值
void get_and_validate_user_input(int num_of_rows){
//...
//Getting the input from the user
scanf("%d",&num_of_rows);
//...
函数参数num_of_rows是函数的局部变量,该变量由传递的参数的值初始化。因此,更改局部变量不会影响原始参数。
您可以像这样声明和定义函数
int get_and_validate_user_input( void )
{
int num_of_rows = 0;
//Input validation
while (1)
{
printf("Enter the number of rows : ");
//Getting the input from the user
scanf("%d",&num_of_rows);
if (num_of_rows <= 0)
{
printf("Negative numbers are not allowed here \n");
}
else
{
break;
}
}
return num_of_rows;
}
并像main一样调用它
num_of_rows = get_and_validate_user_input();