转换IEEE 754浮点数与C十六进制 - 的printf [英] Convert ieee 754 float to hex with c - printf

问题描述

在理想情况下以下code将采取浮动在IEEE 754重新presentation并将其转换为十六进制

Ideally the following code would take a float in IEEE 754 representation and convert it into hexadecimal

void convert() //gets the float input from user and turns it into hexadecimal
{
    float f;
    printf("Enter float: ");
    scanf("%f", &f);
    printf("hex is %x", f);
}

我也不太清楚什么错。它转换成数十六进制数，而是一个非常错误的。

I'm not too sure what's going wrong. It's converting the number into a hexadecimal number, but a very wrong one.

123.1443 gives 40000000
43.3     gives 60000000
8        gives 0

所以它做的事情，我只是不太清楚的东西。

so it's doing something, I'm just not too sure what.

帮助将AP preciated

Help would be appreciated

推荐答案

当你传递一个浮动作为参数传递给一个可变参数函数（如的printf（）），它被提升到一个双击，这是两倍大一个浮动（至少在大多数平台）。

When you pass a float as an argument to a variadic function (like printf()), it is promoted to a double, which is twice as large as a float (at least on most platforms).

要解决这个问题将在浮动强制转换为单向unsigned int类型路过的时候它作为一个参数的printf（）：

One way to get around this would be to cast the float to an unsigned int when passing it as an argument to printf():

printf("hex is %x", *(unsigned int*)&f);

这也比较正确的，因为的printf（）使用格式说明，以确定每个参数有多大。

This is also more correct, since printf() uses the format specifiers to determine how large each argument is.

从技术上讲，该方案违反了严格别名规则。您可以通过复制浮动的字节到 unsigned int类型并传送到<$ C解决这个问题$ C>的printf（）：

Technically, this solution violates the strict aliasing rule. You can get around this by copying the bytes of the float into an unsigned int and then passing that to printf():

unsigned int ui;
memcpy(&ui, &f, sizeof (ui));

printf("hex is %x", ui);

这两种解决方案是基于这样的假设的sizeof（int）的==的sizeof（浮点），这是在许多32位系统的情况下，但ISN 'T一定如此。

Both of these solutions are based on the assumption that sizeof(int) == sizeof(float), which is the case on many 32-bit systems, but isn't necessarily the case.

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