为什么FORTRAN输出数组作为零的元素时，它们是由1/2或1/3相乘？ [英] why is fortran outputting the elements of an array as zeros when they are multiplied by 1/2 or 1/3?

问题描述

所以我写出的folows数组中的元素

So I'm writing out the elements of an array as folows

write(6,'(i4,200(1x,e15.7))')Jtot0, (a*PJjv(i,Jtot0,j,iv),i=1,nenerdif,100)

其中 是一个常量，然而，当这个常数等于 1/2 或 1/3 输出零，如果它等于1，每一件事情顺利的话，数组元素的实际* 8 。
我怎样才能解决这个问题，让我是有义务通过倍的 1/3

where a is a constant, however, when this constant is equal to 1/2 or 1/3 the output is zeros, and if it's equal to 1, every thing goes well, the array elements are real*8. how can i overcome this, giving that i'm obligated to multiply by a factor of 1/3?

推荐答案

在的Fortran 1/2 是一个整数除法操作，这将向下舍入到，在这种情况下， 0 。同为 1/3 。如果你想要一个真实的结果，做一个真正的除法运算，如 1.0 / 2.0 。需要注意的是分配 1/2 的结果到一个真正的变量将真正的变量设置为 0.0 ，即整数除法会导致 0 和分配，这恰好接下来，将这个值转换为与其最接近的实际再presentation。

In Fortran 1/2 is an integer division operation which will round down to, in this case, 0. Same for 1/3. If you want a real result, do a real division operation, such as 1.0/2.0. Note that assigning the result of 1/2 to a real variable will set the real variable to 0.0, that is the integer division will result in 0 and the assignment, which happens next, will cast that value to its nearest real representation.

的整数除法产生整数结果这项业务是很常见的编程语言。

This business of integer division producing integer results is very common in programming languages.

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