C或者C++中,输入一个浮点数,将证书部分和小数部分的数字分别输出,程序应该怎么写?万分感激 编写一个函数,将一个浮点数中的整数部分和小数部分分别输出

\u6c42C/C++ \u7f16\u7a0b \u8f93\u5165\u4e00\u4e2afloat\u578b\u7684\u6570\u636e\uff0c\u5c06\u6574\u6570\u548c\u5c0f\u6570\u90e8\u5206\u5206\u522b\u8f93\u51fa\uff0c\u4f46\u6709\u8981\u6c42\uff01\uff01

\u7ed9\u4f60\u4e2a\u601d\u8def:\u5148\u5c06float\u5f3a\u5236\u8f6c\u6362\u4e3aint,\u5c31\u5f97\u5230\u6b63\u6570\u90e8\u5206\u4e86,\u7136\u540e\u518d\u7528float\u51cf\u53bb\u8fd9\u4e2aint,\u5c31\u5f97\u5230\u5c0f\u6570\u90e8\u5206\u4e86

1\u3001\u7f16\u5199\u51fd\u6570\uff0csp_put_num_2\uff1b\u5185\u5bb9\u4e3a\uff1a
create or replace procedure sp_put_num_2(in_num varchar2,
out_num_1 out varchar2,
out_num_2 out varchar2,
out_info out varchar2) as
n_num number;
begin
begin
n_num := to_number(in_num);
out_num_1 := trunc(n_num);
out_num_2 := case when instr(in_num, '.') > 0 then substr(in_num, instr(in_num, '.') + 1) end;
out_info := '\u8f6c\u6362\u6210\u529f';
exception
when others then
out_info := '\u975e\u6570\u503c\u7c7b\u578b';
end;
end;

2\u3001\u51fd\u6570\u8fdb\u884c\u7f16\u8bd1\uff1b

3\u3001\u8fdb\u884c\u51fd\u6570\u591a\u573a\u666f\u6d4b\u8bd5\uff0c\u5982\u8f93\u5165\u4e3a\u5b57\u7b26\u4e32\u7c7b\uff0c\u5219\u8fd4\u56de\u9519\u8bef\u4fe1\u606f\u2018\u975e\u6570\u503c\u7c7b\u578b\u2019\uff1b

4\u3001\u8f93\u5165\u4e3a\u6570\u5b57\u578b\uff0c\u5982'123.25'\uff0c\u518d\u6b21\u6d4b\u8bd5\uff0c\u5206\u522b\u8fd4\u56de\u6574\u6570\u3001\u5c0f\u6570\u90e8\u5206\uff1b

/*

输入浮点数: 36.98

整数部分:36

小数部分:0.980000

Press any key to continue

*/

#include <stdio.h>

int main() {
double dValue,dDecimals;
int Integer;
printf("输入浮点数: ");
scanf("%lf",&dValue);
Integer = (int)dValue;
dDecimals = dValue - Integer;
printf("整数部分:%d
",Integer);
printf("小数部分:%lf
",dDecimals);
return 0;
}


假定浮点数意味着一个double。

// C
#include <stdio.h>
#include <math.h>
int main(void)
{
    double i, a, b;
    scanf("%lf", &i);
    b = modf(i, &a);
    printf("%f %f", a, b);
}
// C++
#include <iostream>
#include <cmath>
int main() 
{
    double i, a, b;
    std::cin >> i;
    b = std::modf(i, &a);
    std::cout << a << ' ' << b;
}


不需要用库存。。。
#include <stdio.h>
int main(){
float f = 36.98;
int i = f;
float a = f - i;
printf("%d, %f\n", i, a);
}

楼上的那位大哥回答得很好~~~我本想回答的,一看,我还是算了吧

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