问题描述
为什么这个循环（到10亿）只需要几个声音执行...我想 i 是一个32位的整数变量，因此总是小于100亿（大于2 ^ 32），而10亿仍然适合32位的范围（其结束于大约2或4十亿，取决于签名）。虽然我不知道编译器如何提升这个100亿恒定值，但他似乎意识到溢出问题，并使其成为一个无用的循环。
当你使我 a long long int （也许 10000000000 a 10000000000L ，但这似乎没有问题）？
Why does this loop (to 1 billion) only take a few sounds to execute ...
for (i = 0; i < 1000000000; i++) { }... but this loop (to 10 billion) takes >10 minutes?
for (i = 0; i < 10000000000; i++) { }Shouldn't it just take 30 seconds or so (3 seconds x 10)?
解决方案I guess i is a 32-bit integer variable and is therefore always smaller than 10 billion (which is more than 2^32), whereas 1 billion still fits into the 32-bit range (which ends at about 2 or 4 billion, depending on signedness). Though I don't know how the compiler promotes this 10 billion constant, but he seems to realize the overflow issue and makes it an infite loop.
What happens when you make i a long long int (and maybe the 10000000000 a 10000000000L, but that seems to be no problem)?
这篇关于为什么需要更长的时间才能循环100亿比10亿？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！