如何为float的转换/双printf中为int处理？ [英] How is conversion of float/double to int handled in printf?

问题描述

考虑这一计划。

int main()
{
        float f = 11.22;
        double d = 44.55;
        int i,j;

        i = f;         //cast float to int
        j = d;         //cast double to int

        printf("i = %d, j = %d, f = %d, d = %d", i,j,f,d);
        //This prints the following:
        // i = 11, j = 44, f = -536870912, d = 1076261027

        return 0;
}

有人能解释为什么从double /浮到INT铸造正常工作在第一种情况，并在printf的？结果完成后不工作
这个程序被编译在​​GCC-4.1.2的32位Linux机器上。

Can someone explain why the casting from double/float to int works correctly in the first case, and does not work when done in printf?
This program was compiled on gcc-4.1.2 on 32-bit linux machine.

编辑：
<一href=\"http://stackoverflow.com/questions/2398791/how-is-conversion-of-float-double-to-int-handled-in-printf/2398821#2398821\">Zach's回答似乎是合乎逻辑的，即使用格式说明找出弹出堆栈。然而再考虑这个跟进的问题：

Zach's answer seems logical, i.e. use of format specifiers to figure out what to pop off the stack. However then consider this follow up question:

int main()
{

    char c = 'd';    // sizeof c is 1, however sizeof character literal
                     // 'd' is equal to sizeof(int) in ANSI C

    printf("lit = %c, lit = %d , c = %c, c = %d", 'd', 'd', c, c);
    //this prints: lit = d, lit = 100 , c = d, c = 100
    //how does printf here pop off the right number of bytes even when
    //the size represented by format specifiers doesn't actually match 
    //the size of the passed arguments(char(1 byte) & char_literal(4 bytes))    

 return 0;
}

这是如何工作的？

How does this work?

推荐答案

的的printf 函数使用格式说明找出弹出堆栈。所以当它看到％d个，它会弹出4个字节，跨$ P $其中pts他们作为一个 INT ，其中是错误的（二进制重新$ p $ 的psentation（浮点）3.0 是不一样的（INT）3 ）。

The printf function uses the format specifiers to figure out what to pop off the stack. So when it sees %d, it pops off 4 bytes and interprets them as an int, which is wrong (the binary representation of (float)3.0 is not the same as (int)3).

您需要的将是使用％F 格式说明或铸铁的参数 INT 。如果您使用 GCC 新版本不够，那么强烈的警告打开捕获这种错误的：

You'll need to either use the %f format specifiers or cast the arguments to int. If you're using a new enough version of gcc, then turning on stronger warnings catches this sort of error:

$ gcc -Wall -Werror test.c
cc1: warnings being treated as errors
test.c: In function ‘main’:
test.c:10: error: implicit declaration of function ‘printf’
test.c:10: error: incompatible implicit declaration of built-in function ‘printf’
test.c:10: error: format ‘%d’ expects type ‘int’, but argument 4 has type ‘double’
test.c:10: error: format ‘%d’ expects type ‘int’, but argument 5 has type ‘double’

回答这个问题的编辑部分：

Response to the edited part of the question:

C'S整数提升规则说，所有类型的比 INT 较小时得到一个可变参数传递晋升为 INT 。所以你的情况，在D是获得晋升到 INT ，然后printf的是突然离开的 INT 和铸造到字符。我能找到这种行为的最好的参考就是这篇博客。

C's integer promotion rules say that all types smaller than int get promoted to int when passed as a vararg. So in your case, the 'd' is getting promoted to an int, then printf is popping off an int and casting to a char. The best reference I could find for this behavior was this blog entry.

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