问题描述
说我有一个C程序，在pseudoish是：
Say I have a C program that in pseudoish is:
For i=0 to 10
x++
a=2+x*5
next
时触发器为这个数字（1 [X ++] + 1 [X * 5 + 1 [2 +（X + 5））] * 10 [循环]，30 FLOPS？我无法理解一个翻牌是什么。
Is the number of FLOPs for this (1 [x++] + 1 [x*5] + 1 [2+(x+5))] * 10[loop], for 30 FLOPS? I am having trouble understanding what a flop is.
注意[...]都表明在那里我从让我计数操作。
Note the [...] are indicating where I am getting my counts for "operations" from.
推荐答案
有关FLOPS测量的目的，通常只有加法和乘法都包括在内。像师，倒数，平方根和超越函数东西都很贵，包括作为一个单一的操作，而像载入和存储是太微不足道了。
For the purposes of FLOPS measurements, usually only additions and multiplications are included. Things like divisions, reciprocals, square roots, and transcendental functions are too expensive to include as a single operation, while things like loads and stores are too trivial.
在换句话说，你的循环体内含有2加1乘，所以（假设 X 是浮点）每次循环迭代为3 OPS;如果运行循环10次，你已经做了30欢声笑语。
In other words, your loop body contains 2 adds and 1 multiply, so (assuming x is floating point) each loop iteration is 3 ops; if you run the loop 10 times you've done 30 ops.
请注意，衡量MIPS的时候，你的循环将超过3指令，因为它也包括加载和存储的FLOPS测量不计数。
Note that when measuring MIPS, your loop would be more than 3 instructions because it also includes loads and stores that the FLOPS measurement doesn't count.
这篇关于什么是一个失败计数？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！