为什么要分两个整数没有得到一个浮动？ [英] Why dividing two integers doesn't get a float?

问题描述

任何人都可以解释为什么B获得在这里圆满结束时，我用一个整数除以它，虽然它是一个浮？

Can anyone explain why b gets rounded off here when I divide it by an integer although it's a float?

#include <stdio.h>

void main() {
    int a;
    float b, c, d;
    a = 750;
    b = a / 350;
    c = 750;
    d = c / 350;
    printf("%.2f %.2f", b, d);
    // output: 2.00 2.14
}

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推荐答案

这是因为隐式转换。变量 B，C，D 是浮动键入。但 / 操作员可以看到两个整数它来划分，因此在被隐式转换为浮动通过加入小数点。如果你想浮动师，尝试使两个操作数的 / 浮动。就像如下：

This is because of implicit conversion. The variables b, c, d are of float type. But the / operator sees two integers it has to divide and hence returns an integer in the result which gets implicitly converted to a float by the addition of a decimal point. If you want float divisions, try making the two operands to the / floats. Like follows.

#include <stdio.h>

int main() {
    int a;
    float b, c, d;
    a = 750;
    b = a / 350.0f;
    c = 750;
    d = c / 350;
    printf("%.2f %.2f", b, d);
    // output: 2.14 2.14
    return 0;
}

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