Floating point 将原始二进制文件转换为小值的浮点打印0
我试图读取2个32位值（假设它们是ieee 745单精度浮点值），并使用以下代码段比较它们以检查它们是否相等Floating point 将原始二进制文件转换为小值的浮点打印0,floating-point,intel,floating-point-precision,floating-point-conversion,Floating Point,Intel,Floating Point Precision,Floating Point Conversion,我试图读取2个32位值（假设它们是ieee 745单精度浮点值），并使用以下代码段比较它们以检查它们是否相等 #include <stdio.h> #include <stdint.h> void main(){ int a; int b; //a = 0x3f99999a ; //b = 0x3fa66666 ; a = 0xfa98 ; b = 0x65cc ; printf("Comparison is %d
#include <stdio.h>
#include <stdint.h>
void main(){
int a;
int b;
//a = 0x3f99999a ;
//b = 0x3fa66666 ;
a = 0xfa98 ;
b = 0x65cc ;
printf("Comparison is %d\n",((*(float*)&(a))==(*(float*)&(b))));
printf("Numbers are a= %f and b = %f \n",(*(float*)&(a)),(*(float*)&(b)));
}
Comparison is 0 // This is correct
Numbers are a= 1.2 and b = 1.3 // True decimal values for given 745 representation
Comparison is 0 // This is correct
Numbers are a= 1.2 and b = 1.3 // True decimal values for given 745 representation
Comparison is 1 // This is clearly wrong as both numbers are unequal
Numbers are a= 0 and b = 0 // Casting it to float is causing them to 0 ??
Comparison is 0 // This is correct !!
Numbers are a= 0 and b = 0 // Still 0 because numbers are subnormal
Comparison is 0 // This is correct
Numbers are a= 1.2 and b = 1.3 // True decimal values for given 745 representation
Comparison is 0 // This is correct
Numbers are a= 1.2 and b = 1.3 // True decimal values for given 745 representation
所以我的问题是为什么平台A产生了错误的浮动比较结果。第一组值低于正常值，在十进制表示中非常小，但绝对不相等。其中，as代码适用于第二组值（大值）。谁能给我解释一下吗。为什么平台A不显示值与平台B不相同 问题是mxcsr寄存器的DAZ位。一台机器A设置了它，所以它将非规范化处理为0。其中，在机器2上，DAZ被重置，因此它正在对其进行规范化，并将非规范化视为不同的值，这导致比较结果不相等。但我真的很感激所有帮助我的人：）！没有你的帮助是不可能调试的 请注意，
0xfa980x65cc%f%e数字是a=8.989610e-41和b=3.651784e-41