英文:
C Code to convert from sign magnitude to two's complement using restricted operators
问题
我正在用C语言编写代码，将原码转换为二进制补码。最高位是符号位，输入将是short类型。到目前为止，我有一个代码，但测试未通过（我没有测试案例）：
int magtoComp(int x) {
int mask = x >> 15;
int mag = x & ~(1 << 15);
return (mag ^ mask) + ~mask + 1;
英文:
I am writing a code in C to convert from sign magnitude to two's compliment. The most significant bit is the sign bit and the input will be a short type.
thus far I have a code that is failing the test (I do not have test cases):
int magtoComp(int x) {
int mask = x >> 15;
int mag = x & ~(1 << 15);
return (mag ^ mask) + ~mask + 1;
答案1
得分: 3
计算二进制补码表示中的 -x，可以执行 ~x + 1。如果 sign 是符号位，可以写成 (x ^ -sign) + sign。- 不被允许，但可以计算 -sign 为 ~sign + 1。
因此，这个解决方案：
// 将一个16位符号-幅度表示转换为二进制补码
int shortSignMag2TwosComp(int x) {
unsigned y = x; // 使用无符号算术
unsigned sign = y >> 15; // 提取符号位
unsigned mag = y & ~(1u << 15); // 屏蔽掉符号位
return (mag ^ (~sign + 1)) + sign;
}
如果 int 有超过16位，你可以使用 int 替代 unsigned 来符合规则：
// 将一个16位符号-幅度表示转换为二进制补码
// 假设 int 类型具有超过15位值位
int shortSignMag2TwosComp(int x) {
int sign = x >> 15; // 提取符号位
int mag = x & ~(1 << 15); // 屏蔽掉符号位
return (mag ^ (~sign + 1)) + sign;
}
英文:
To compute -x in two's complement representation, one performs ~x + 1. if sign is the sign bit, this can be written (x ^ -sign) + sign. - is not allowed but you can compute -sign as ~sign + 1.
Hence this solution:
// convert a 16-bit sign+magnitude representation to 2's complement
int shortSignMag2TwosComp(int x) {
unsigned y = x; // use unsigned arithmetics
unsigned sign = y >> 15; // extract the sign bit
unsigned mag = y & ~(1u << 15); // mask off the sign bit
return (mag ^ (~sign + 1)) + sign;
}
If int has more than 16 bits, you can use int in place of unsigned to conform to the rules:
// convert a 16-bit sign+magnitude representation to 2's complement
// assuming type int has more than 15 value bits
int shortSignMag2TwosComp(int x) {
int sign = x >> 15; // extract the sign bit
int mag = x & ~(1 << 15); // mask off the sign bit
return (mag ^ (~sign + 1)) + sign;
}
答案2
得分: 0
假设你的计算机是2的补码形式，这就很简单：
uint32_t sign = data & (1u << 31);
if(sign)
{
data &= ~(1u << 31);
data *= -1;
}
或者，稍微优化一些的版本：
int32_t signmag_to_2compl (uint32_t signmag)
{
int32_t sign = (signmag & (1u << 31)) ? -1 : 1;
signmag &= ~(1u << 31);
return sign * signmag;
}
`uint32_t`，因为在2的补码机器上运行时，将有符号数存储为`int`没有任何意义。我们还希望避免在有符号数上进行位运算。
英文:
Assuming your computer is 2's complement, it's as easy as:
uint32_t sign = data & (1u << 31);
if(sign)
{
data &= ~(1u << 31);
data *= -1;
}
Or if you will, a slightly more optimized version:
int32_t signmag_to_2compl (uint32_t signmag)
{
int32_t sign = (signmag & (1u << 31)) ? -1 : 1;
signmag &= ~(1u << 31);
return sign * signmag;
}
uint32_t since it's completely senseless to store a signed magnitude number in an int while running on a 2's complement machine. We also want to avoid doing bitwise arithmetic on signed numbers.
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