我写了一个决定三角形类型的程序。当用户输入字符串输入时,它会给出错误。我想为双输入提供错误。为了做到这一点,我尝试了这个
else if (s1[i] == '.') {
found_double = 1;
break;
但程序也将.
识别为字符串。我怎么解决这个问题?完整代码如下。
/*
* HW3-3.c
*
* Created on: Oct 21, 2012
* Author: mert
*/
#include <string.h>
#include <stdio.h>
void checkTriangle(char *s1,char *s2,char *s3)
{
int i;
int found_double = 0;
int found_letter = 0;
int len = strlen(s1);
int len2 = strlen(s2);
int len3 = strlen(s3);
for( i = 0; i < len; i++)
{
if(s1[i] < '0' || s1[i] > '9')
{
found_letter = 1; // this variable works as a boolean
break;
} else if (s1[i] == '.') {
found_double = 1;
break;
}
}
for( i = 0; i < len2; i++)
{
if(s2[i] < '0' || s2[i] > '9')
{
found_letter = 1; // this variable works as a boolean
break;
} else if (s2[i] == '.') {
found_double = 1;
break;
}
}
for( i = 0; i < len3; i++)
{
if(s3[i] < '0' || s3[i] > '9')
{
found_letter = 1; // this variable works as a boolean
break;
} else if (s3[i] == '.') {
found_double = 1;
break;
}
}
if(found_letter) // value 0 means false, any other value means true
printf("Please enter an integer instead of string.");
else if (found_double)
printf("Please enter an integer instead of double.");
else
{
int side1 = atoi(s1);
int side2 = atoi(s2);
int side3 = atoi(s3);
if ((side1 + side2 > side3 && side1 + side3 > side2 && side2 + side3 > side1) && (side1 > 0 && side2 > 0 && side3 > 0))
{
// Deciding type of triangle according to given input.
if (side1 == side2 && side2 == side3)
printf("EQUILATERAL TRIANGLE");
else if (side1 == side2 || side2 == side3 || side1 == side3)
printf("ISOSCELES TRIANGLE\n");
else
printf("SCALENE TRIANGLE \n");
}
else
printf("\nTriangle could not be formed.");
}
}
int main(void)
{
char s[32], s2[32], s3[32];
printf("Please enter sides of triangle");
printf("\nPlease enter side 1:");
gets(s);
printf("Please enter side 2:");
gets(s2);
printf("Please enter side 3:");
gets(s3);
checkTriangle(s,s2,s3);
}
答案 0 :(得分:0)
答案 1 :(得分:0)
您可以更改IF的顺序。
if (s1[i] == '.')
{
//code here
}
else if(s1[i] < '0' || s1[i] > '9')
{
//code here
}
因为'。'的ascii代码定义为不在区间['0','9']中,所以它总是会被捕获在第一个IF中。
您可以尝试以下代码来查看其ascii值:
printf("Ascii code for char '.' is %d\n", '.');
printf("Ascii code for char '0' is %d\n", '0');
printf("Ascii code for char '9' is %d\n", '9');