的printf（QUOT;％F＆QUOT ;, AA）时，AA是int类型 [英] printf("%f", aa) when aa is of type int

问题描述

double a = 3.1456;

int aa=2.0;      

printf ("%f \n",a );    
printf("%f \n",aa);

答案是： 3.145600，3.145599。
我已经尝试了不同的值，并且输出AA似乎有一定的关系。
我很困惑。什么原因？
〜结果
〜结果
〜结果
〜

The answer is 3.145600 ,3.145599. I have tried different values of a, and the outputs aa seem to have some relationship with a. I am confused. What's the reason? ~
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~
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推荐答案

英戈伦哈德是正确地说，你的程序行为是不确定的 - 输出不会成为下优化或在不同的架构/编译器保持一致。

Ingo Leonhardt is correct to say that your program's behaviour is undefined - the output isn't going to be consistent under optimisation or on different architectures / compilers.

但是，你看到3.1456在这种情况下，突变为3.145599的具体原因是可解释的。由此可以推断，你的架构采用IEEE-754 8个字节的双击和4个字节的小端 INT 秒。最近的8字节双至3.1456的十六进制再presentation是：

However, the specific reason that you're seeing 3.1456 mutated to 3.145599 in this case is explicable. It can be inferred that your architecture uses IEEE-754 8-byte doubles and 4-byte little-endian ints. The hexadecimal representation of the nearest 8-byte double to 3.1456 is:

0x40092a305532617c

，其中的符号 0 ，指数字段是 0x400的（重presenting 1指数）和尾数字段是 0x92a305532617c （重presenting 0x1.92a305532617c十六进制尾数）。如果覆盖的4字节整数2的低4个字节的这一点，那么你会得到：

where the sign is 0, the exponent field is 0x400 (representing an exponent of 1) and the mantissa field is 0x92a305532617c (representing a hexadecimal mantissa of 0x1.92a305532617c). If you overwrite the lower 4 bytes of this with the 4-byte integer 2, then you get:

0x40092a3000000002

这具有符号和指数不变，但尾数变为0x1.92a3000000002，或比原始0x0.000005532617a以下。随着1的指数，这个重新presents的0x0.00000aa64c2f4差异，这在小数为〜0.000000634766。当您从3.1456减去这个你〜3.145599365234，这轮以3.145599。

This has the sign and exponent unchanged, but the mantissa has changed to 0x1.92a3000000002, or 0x0.000005532617a less than the original. With an exponent of 1, this represents a difference of 0x0.00000aa64c2f4, which in decimal is ~0.000000634766. When you subtract this from 3.1456 you get ~3.145599365234, which rounds to 3.145599.

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