转换上浮到只使用位操作一个int(float2int) [英] Converting float to an int (float2int) using only bitwise manipulation
问题描述
我想知道,如果有人可以让我与我的工作有问题的正确方向。我想下面的C函数不使用的是什么只有ARM汇编以及位操作要做到:
INT float2int(浮X){
回报(INT)X;
}
我已经codeD这个(int2float),而不许多问题相反。我只是不能确定在哪里这个新的问题开始。
例如:
3(INT)= 0x40400000(浮点)
0011 = 0千万10000000000000000000000
其中0为符号位,千万是指数,而10000000000000000000000是尾数/分。
有人能简单地指出我这个问题的正确方向?即使是一个C伪code再presentation将是有益的。我知道我需要提取符号位,提取指数和反向偏压(127),并提取部分,但我只是不知道从哪里开始。
还有,如果浮子不能重新presented为整数的问题(因为它溢出或为NaN)。
任何帮助将是AP preciated!
//假设int可以容纳所有的浮点的precision。
INT float2int(浮X){
INT注册= f_SignRawBit(X);
无符号的尾数为f_RawMantissaBits(X); // 0 - 0x7FFFFF
INT世博= f_RawExpoBits(X); // 0 - 255
//形成正确的指数和尾数
如果(世博== EXPO_MAX){
Handle_NAN_INF();
}
否则,如果(世博== EXPO_MIN){
世博+ = BIAS + 1 - MantissaOffset / * 23 * /;
}
其他{
世博+ = BIAS - MantissaOffset / * 23 * /;
尾数| = ImpliedBit;
}
而(世博会大于0){
Expo--;
//添加code检测溢出
尾数* = 2;
}
而(博览会暨小于0){
世博++;
//添加code要注意最后移出位
//添加code注意如果有非零位移出
尾数/ = 2;
} //添加四舍五入code取决于`最后一个移出bit`和'非零位移位out`。如果向0四舍五入可能不被需要的。 //添加code检测过/欠流在以下
如果(登录){
返回-Mantissa;
}
返回尾数;
}
I am wondering if someone could set me in the right direction with a problem I am working on. I am trying to do what the following C function does using only ARM assembly and bit manipulation:
int float2int(float x) {
return (int) x;
}
I have already coded the reverse of this (int2float) without many issues. Im just unsure of where to start with this new problem.
For example:
3 (int) = 0x40400000 (float)
0011 = 0 10000000 10000000000000000000000
Where 0 is the Sign Bit, 10000000 is the exponent, and 10000000000000000000000 is the mantissa/fraction.
Can someone simply point me in the right direction with this problem? Even a C pseudocode representation would be helpful. I know I need to extract the sign bit, extract the exponent and reverse the bias (127) and also extract the fraction but I just have no idea where to begin.
There is also the issue of if the float cannot be represented as an integer (because it overflows or is a NaN).
Any help would be appreciated!
// Assume int can hold all the precision of a float.
int float2int(float x) {
int Sign = f_SignRawBit(x);
unsigned Mantissa = f_RawMantissaBits(x); // 0 - 0x7FFFFF
int Expo = f_RawExpoBits(x); // 0 - 255
// Form correct exponent and mantissa
if (Expo == EXPO_MAX) {
Handle_NAN_INF();
}
else if (Expo == EXPO_MIN) {
Expo += BIAS + 1 - MantissaOffset /* 23 */;
}
else {
Expo += BIAS - MantissaOffset /* 23 */;
Mantissa |= ImpliedBit;
}
while (Expo > 0) {
Expo--;
// Add code to detect overflow
Mantissa *= 2;
}
while (Expo < 0) {
Expo++;
// Add code to note last shifted out bit
// Add code to note if any non-zero bit shifted out
Mantissa /= 2;
}
// Add rounding code depending on `last shifted out bit` and `non-zero bit shifted out`. May not be need if rounding toward 0.
// Add code to detect over/under flow in the following
if (Sign) {
return -Mantissa;
}
return Mantissa;
}
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