我正在做作业，但被卡住了。由于某些原因，我无法获得以下结果：
byte order: little-endian
> FFFFFFFF
0xFFFFFFFF
signBit 1, expBits 255, fractBits 0x007FFFFF
QNaN
> 3
0x00000003
signBit 0, expBits 0, fractBits 0x00000003
denormalized: exp = -126
> DEADBEEF
0xDEADBEEF
signBit 1, expBits 189, fractBits 0x002DBEEF
normalized: exp = 62
> deadbeef
0xDEADBEEF
signBit 1, expBits 189, fractBits 0x002DBEEF
normalized: exp = 62
> 0
0x00000000
signBit 0, expBits 0, fractBits 0x00000000
+zero
我曾尝试解决此问题，但我不知道哪里出了问题。以下代码是我对该项目的尝试。我觉得我快结束了，但我还没完。
这是我的代码：
#include <stdio.h>
#include <stdlib.h>
static char *studentName = "name";
// report whether machine is big or small endian
void bigOrSmallEndian()
{
int num = 1;
if(*(char *)&num == 1)
{
printf("\nbyte order: little-endian\n\n");
}
else
{
printf("\nbyte order: big-endian\n\n");
}
}
// get next int (entered in hex) using scanf()
// returns 1 (success) or 0 (failure)
// if call succeeded, return int value via i pointer
int getNextHexInt(int *i)
{
// replace this code with the call to scanf()
//*i = 0;
//return 1;
scanf ("%x", i);
return 1;
}
// print requested data for the given number
void printNumberData(int i)
{
//printf("%x %0#10x\n",i,*(int *)&i);
int tru_exp =0;
//int stored_exp;
int negative;
int exponent;
int mantissa;
printf("\n>");
printf("\n0x%08X",i);
negative = !!(i & 0x80000000);
exponent = (i & 0x7f800000) >> 23;
mantissa = (i & 0x007FFFFF);
printf("\nsignBit %d, ", negative);
printf("expbits %d, ", exponent);
printf("fractbits 0x%08X", mantissa);
// "%#010x, ", mantissa);
if(exponent == 0)
{
if(mantissa != 0)
{
printf("\ndenormalized ");
}
}
else{
printf("\nnormalized: ");
tru_exp = exponent - 127;
printf("exp = %d", tru_exp);
}
if(exponent == 0 && mantissa == 0 && negative == 1)
{
printf("\n-zero");
}
if(exponent ==0 && mantissa == 0 && negative == 0)
{
printf("\n+zero");
}
if(exponent == 255 && mantissa != 0 && negative == 1)
{
printf("\nQNaN");
}
if(exponent == 255 && mantissa != 0 && negative == 0)
{
printf("\nSNaN");
}
if(exponent == 0xff && mantissa == 0 && negative == 1)
{
printf("\n-infinity");
}
if(exponent == 0xff && mantissa == 0 && negative == 0)
{
printf("\n+infinity");
}
printf("\n");
while(i != 0)
break;
}
// do not change this function in any way
int main(int argc, char **argv)
{
int i; // number currently being analyzed
int nValues; // number of values successfully parsed by scanf
printf("CS201 - A01p - %s\n\n", studentName);
bigOrSmallEndian();
for (;;) {
if (argc == 1) // allow grading script to control ...
printf("> "); // ... whether prompt character is printed
nValues = getNextHexInt(&i);
printf("0x%08X\n", i);
if (! nValues) { // encountered bad input
printf("bad input\n");
while (getchar() != '\n') ; // flush bad line from input buffer
continue;
}
printNumberData(i);
if (i == 0)
break;
}
printf("\n");
return 0;
}
此代码的结果是：
byte order: little-endian
> FFFFFFFF
0xFFFFFFFF
>
0xFFFFFFFF
signBit 1, expbits 255, fractbits 0x007FFFFF
normalized: exp = 128
QNaN
> 3
0x00000003
>
0x00000003
signBit 0, expbits 0, fractbits 0x00000003
denormalized
> DEADBEEF
0xDEADBEEF
>
0xDEADBEEF
signBit 1, expbits 189, fractbits 0x002DBEEF
normalized: exp = 62
> deadbeef
0xDEADBEEF
>
0xDEADBEEF
signBit 1, expbits 189, fractbits 0x002DBEEF
normalized: exp = 62
> 0
0x00000000
>
0x00000000
signBit 0, expbits 0, fractbits 0x00000000
+zero
我相信问题出在
printNumberData()，但我无法指出问题所在。我希望有人能参与进来，因为我们将不胜感激。
最佳答案
因此，据我所知，唯一的问题是您的指数未签名。您可以通过将tru_exp更改为int8_t（或signed char）来解决此问题。或者通过简单地手动对数字进行符号扩展，例如if (tru_exp > 128) tru_exp -= 256;
关于c - 分析IEEE 754位模式，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/14797418/