与C中的sqrt（）函数混淆

Question

When i do like this, it works well in my code...

...

for (i = 2; i <= sqrt(500000); i++)

...

But do like

for (i = 2; i < sqrt(500000) + 1; i++)

executes after compile, error occurs saying Segmentation fault (core dumped) .

for loop body is:

for (i = 2; i <= sqrt(500000); i++) {
        summation[i * i] += i;
        for (j = i + 1; j <= 500000 / i; j++) {
            summation[i * j] += (i + j);
        }
    }

Is there any difference between the two for loops? Thanks

Answer 1

你的第二个循环再次运行：500000不是一个完美的正方形，所以i < sqrt(500000)和i <= sqrt(500000)总是相等，+1确保另一次迭代。

Answer 2

sqrt()将返回一个浮点数，因此所有比较都是针对浮点数进行的，并且它们在设计上是不精确的，因此上面的这两个比较可以产生不同的结果 ，因此在一个案例中你将得到一个 -一个和未定义的行为。

Answer 3

I've tried your code with gcc 4.3.4 under cygwin.

• In the first case, i loops up to 707.
• In the second case, i loops up to 708.

I bet that this last value triggers a buffer overflow somewhere in the body of the loop.

Answer 4

As others have already explained, the + 1 causes the loop to iterate once too often. Then summation[i * i] or summation[i * j] accesses beyond the allocated size of summation . The solution is to either increase the allocated size accordingly or make sure that the condition is correct (not doing + 1 ) and you thus do not run over the end of the array.

But also, like others already said, you should not use a floating point value (result of sqrt ) with an integer comparison as floating point values are... tricky. I'm not sure whether in this case the int gets casted to float or vice versa, but whichever way it's not the right thing to do.

Answer 5

did you try this?

for (i = 2; i < (sqrt(500000) + 1); i++)

Is your ia floating pont variable?