c++中如何检查输入的是否为数字,谢谢

街头浪者

c++中如何检查输入的是否为数字,谢谢

玉儿

寒，你按字符串读入，然后-个‘0’就知道了么

绝对标致

定义一个数据类型为 int
然后对 cin 做是否成功判断。

熾天使rose

输入的应该是字符串 然后强制INT转换成INT型的 再进行判断 也行否？

Forever紫月

#include <iostream.h>
#include <string.h>
#include <ctype.h>

bool IsNumber( const char *number ) {
bool isnumber = true;
int len = strlen( number );
if(len == 0 || len > 9) {
return !isnumber;
}
for( int i = 0; i < len; i++ )
{
if( isdigit( number[i] ) == 0 )
{
if( number[i] == '.' )
{
if( i + 1 > len - 1 )
{
isnumber = false;
break;
}
}
else if( number[i] == '+' || number[i] == '-' )
{
if( i - 1 >= 0 )
{
isnumber = false;
break;
}
}
else
{
isnumber = false;
break;
}
}
}
return isnumber;
}

void main()
{
char number[20] = {0};
cout << "Enter a number: ";
cin >> number;
if(!IsNumber(number))
cout << "This is not a valid number" << endl;
else
cout << "Thanks for entering a number!" << endl;
}

qz沒愛過mm

直接用getch();
判断它的ＡＳＣＩＩ的值就行拉
getch() 左支有个功能是　如果是
int x;
x=getch();这样x里面是这个字母的ＡＳＣＩＩ码
用这个判断就行．

林枫

#include <iostream.h>
#include <string.h>
#include <ctype.h>
#define NUM "0123456789."
bool IsNumber( const char *number ) {
bool isnumber = true;

int len = strlen( number );
if(len == 0 || len > 9) {
return !isnumber;
}

if (number[0] == '-' || number[0] == '+') {
for(int i=0; i<len -1; ++i ) {
if(strchr(NUM, number[i + 1]) == NULL)
return !isnumber;
}

}else{
for(int i=0; i<len; ++i ) {
if(strchr(NUM, number[i]) == NULL)
return !isnumber;
}
}

return isnumber;
}
void main()
{
char number[20] = {0};
cout << "Enter a number: ";
cin >> number;
while (!IsNumber(number)) {
cout << "not number" << endl;
cout << "Enter a number: ";
cin >> number;
}
cout << "OK!" << endl;
}