请教：如何判断一个浮点型变量的值是不是整数？

[TinkyWinky]

哪位大虾能告诉我判断一个浮点型变量的值是不是整数的最简单方法？

[星羽]

#include <stdio.h>


int main()
{
    float x = 0.f;
    scanf("%f", &x);

    if (x == (int)x)
        printf("is int\n");

    return 0;

}

[星羽]

#include <stdio.h>


int main()
{
    double x = 0.f;

    while (scanf("%lf", &x) == 1) {
        if (x == (int)x)
            printf("is int\n");
        else
            printf("is not int\n");
    }
    return 0;

}


在double精度内有效


--------

测试

1
is int
2
is int
1.1
is not int
1.00000000001
is not int
1.0000123123000123
is not int
12323123.000
is int
a

[xiaoc10]

#include <stdio.h>
#include <iostream>
using namespace std;

int main()
{
    float result = 0;
    int x;
    char input[100],c = 'n';
    while( scanf("%s", input) )
    {
        sscanf(input, "%d%c", &x, &c);
        if(c == '.')
        {
            cout << input << " is not a int." << endl;
            sscanf(input, "%f", &result);
            cout << "The input is set to result: " << result << endl;
            c = 'n';
        }
        else
            cout << input  << " is a int" << endl;
    }
    system("pause");
    return 0;
}

[svchost]

楼上用的什么编译器？

[kojie_chen]

x==(int)x
或者 就是判断小数点了

[TinkyWinky]

[Quote=引用 1 楼 Chiyer 的回复:]
C/C++ code
#include <stdio.h>


int main()
{
float x = 0.f;
scanf("%f", &x);

if (x == (int)x)
printf("is int\n");

return 0;

}
[/Quote]
可能是我说得不清楚。。
补充一下：
如果那个浮点型变量的值是从cin中输入的话应该怎么判断呢？
比如我输入1.0它就判断是浮点型，输入1它就判断为整型。

[星羽]

随便写了下，你参考下吧

#include "iostream"
using namespace std;

enum {efloat, eint, notnum };

int check(const char str[]) {
    
    int dot = 0;

    for (size_t i = 0; i < strlen(str); ++i) {

        if (!isdigit(str[i])) {

            if (str[i] == '.') {
                ++dot;
            if (dot > 1)
                return notnum;
            } else {
                return notnum;
            }

        }
    }

    if (dot == 1)
        return efloat;

    return eint;
}

int main()
{
    double d = 0.0;
    char t[128];
    
    bool bcontinue = true;

    while (bcontinue) {

        cin>>t;

        switch (check(t)) {
            case efloat :
                cout<<"is float"<<endl;
                break;
            case eint :
                cout<<"is int"<<endl;
                break;
            default :
                bcontinue = false;
                break;
                
        }



    }
    
    return 0;

}


--------

测试

1
is int
123123
is int
1.123123
is float
1.00000000000000000000001
is float
1.0
is float
1
is int
a

[c_spark]

如含有非数字字符，则不是整数

#include <stdio.h>

int main()
{
	char str[20];
	int flag = 0;
	int v;
	flag = 0;
	scanf("%s",str);
	for(int i = 0; str[i] != '\0'; ++i)
	{
		if(!(str[i] >= '0'&&str[i] <= '9'))
		{
			flag = 1;
			break;
		}
	}
	if(flag == 0)
	{
		sscanf(str,"%d",&v);
		printf("%d is integer\n",v);
	}
	else
		printf("%s is not integer",str);
	return 0;
}

1.0
1.0 is not integer

[TinkyWinky]

楼上两位高人都是用字符数组来获取输入的值，而我是想让一个浮点型变量获取输入的值。。。。

[星羽]

#include "iostream"
using namespace std;

enum {efloat, eint, notnum };

int check(const char str[]) {

    int dot = 0;

    for (size_t i = 0; i < strlen(str); ++i) {

        if (!isdigit(str[i])) {

            if (str[i] == '.') {
                ++dot;
                if (dot > 1)
                    return notnum;
            } else {
                return notnum;
            }

        }
    }

    if (dot == 1)
        return efloat;

    return eint;
}

int main()
{
    double d = 0.0;
    int n = 0;
    char t[128];

    bool bcontinue = true;

    while (bcontinue) {

        cin>>t;

        switch (check(t)) {
            case efloat :
                d = atof(t);        // 转成double
                cout<<"is float "<<d<<endl;
                break;
            case eint :
                n = atoi(t);       // 转成int
                cout<<"is int "<<n<<endl;
                break;
            default :
                bcontinue = false;
                break;

        }



    }

    return 0;

}