int 不可能进行二进制移位
usigned int val = 1;
val <<= 30;
cout << intToBin(val) << endl;
string intToBin(unsigned int val) {
unsigned int k=1;
string ret;
while (k <= val) {
if (k & val) {
ret.insert(0,"1");
} else {
ret.insert(0,"0");
}
k <<= 1;
}
return ret;
}
这将写入 1 和 30x 0，这是可以的。但我需要的是在最高位上有 1，这意味着在第一个位置 - 后面跟着 31 个零。但是当我尝试 val <<= 31;什么也没写，我看不懂。您能为我澄清一下吗？
谢谢
usigned int val = 1;
val <<= 30;
cout << intToBin(val) << endl;
string intToBin(unsigned int val) {
unsigned int k=1;
string ret;
while (k <= val) {
if (k & val) {
ret.insert(0,"1");
} else {
ret.insert(0,"0");
}
k <<= 1;
}
return ret;
}
This will write 1 and 30x 0 which is alright. But what I need is to have the 1 on the highest bit, meaning on the first position - followed by 31x zeros. But when I try to val <<= 31; nothing is written, which I don't understand. Could you clarify that for me please?
Thank you
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如果
val为>= 2^31，您的 while 循环将不会终止。这是因为
k == 2^31仍然是<=val但2^31 << 1 == 2^32溢出并变为 0。这仍然小于限制。您也可以扩展您的条件以中断 if
k == 0，那么问题应该消失。（本文中
^表示求幂而不是异或）Your while loop will not terminate if
valis>= 2^31.This is because
k == 2^31is still<=valbut2^31 << 1 == 2^32overflows and becomes 0. Which is still smaller than the limit.You could extend your condition to break if
k == 0too, then the problem should disappear.(
^means exponentiation not xor in this post)