比较浮点型和双精度型
#include <stdio.h>
int main(void){
float a = 1.1;
double b = 1.1;
if(a == b){
printf("if block");
}
else{
printf("else block");
}
return 0;
}
打印：else block
#include <stdio.h>
int main(void){
float a = 1.5;
double b = 1.5;
if(a == b){
printf("if block");
}
else{
printf("else block");
}
return 0;
}
打印：if block
这背后的逻辑是什么？
使用的编译器：gcc-4.3.4
#include <stdio.h>
int main(void){
float a = 1.1;
double b = 1.1;
if(a == b){
printf("if block");
}
else{
printf("else block");
}
return 0;
}
Prints: else block
#include <stdio.h>
int main(void){
float a = 1.5;
double b = 1.5;
if(a == b){
printf("if block");
}
else{
printf("else block");
}
return 0;
}
Prints: if block
What is the logic behind this?
Compiler used: gcc-4.3.4
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评论（3）
这是因为
1.1不能完全用二进制浮点表示。但1.5是。因此，
float和double表示形式将保存稍微不同的1.1值。这正是写为二进制浮点时的区别：
因此，当您比较它们时（并且
float版本得到提升），它们将不相等。This is because
1.1is not exactly representable in binary floating-point. But1.5is.As a result, the
floatanddoublerepresentations will hold slightly different values of1.1.Here is exactly the difference when written out as binary floating-point:
Thus, when you compare them (and the
floatversion gets promoted), they will not be equal.必读：每个计算机科学家应该了解的浮点运算< /a>
Must read: What Every Computer Scientist Should Know About Floating-Point Arithmetic
二进制 1.1 十进制的精确值是非结尾分数 1.00011001100110011001100(1100).... 双常量
1.1是尾数的 53 位截断/近似值。现在，当转换为浮点数时，尾数将仅以 24 位表示。当
float转换回 double 时，尾数现在又回到 53 位，但超过 24 位的所有内存都将丢失 - 该值被零扩展，现在您正在比较 (例如，取决于舍入行为）并且
现在，如果您使用 1.5 而不是 1.1；
十进制 1.5 恰好是二进制的 1.1。它可以精确地用2位尾数表示，因此即使是24位浮点数也是夸张的......你所拥有的是
和
后者，零扩展到双精度将是
这显然是相同的号码。
The exact value of 1.1 decimal in binary is non-ending fraction 1.00011001100110011001100(1100).... The double constant
1.1is 53-bit truncation / approximate value of that mantissa. Now this when converted to float, the mantissa will be represented just in 24 bits.When the
floatis converted back to double, the mantissa is now back to 53 bits, but all memory of the digits beyond 24 are lost - the value is zero-extended, and now you're comparing (for example, depending on the rounding behaviour)and
Now, if you used 1.5 instead of 1.1;
1.5 decimal is exactly 1.1 in binary. It can be presented exactly in just 2 bit mantissa, therefore even the 24 bits of float are an exaggeration... what you have is
and
The latter, zero extended to a double would be
which clearly is the same number.