division

Division in a SQL statement.

那年仲夏 提交于 2019-12-03 17:28:44
Ok so my question: I have a table itemconfig where there is lots of data concerning items stored in our warehouses. I need to select a special group of items so I can do some job related testing. So far I've been doing the math in my head as I scroll through the database but there must be an easier way. Within itemconfig I want to specifically look at columns case_qty and pal_qty and itm_num. What I would like to do is select all itm_num where pal_qty / case_qty is greater than say 500 . This would give me all itm_num instantly that are relevant to my tests. Sadly I'm not familiar with how to

Why is 0 divided by 0 an error?

那年仲夏 提交于 2019-12-03 11:04:24
问题 I have come across this problem in a calculation I do in my code, where the divisor is 0 if the divident is 0 too. In my code I return 0 for that case. I am wondering, while division by zero is generally undefined, why not make an exception for this case? My understanding why division by zero is undefined is basically that it cannot be reversed. However, I do not see this problem in the case 0/0. EDIT OK, so this question spawned a lot of discussion. I made the mistake of over-eagerly

Does INT_MIN % -1 produce undefined behavior?

馋奶兔 提交于 2019-12-03 09:37:45
gcc generates floating code that raises SIGFPE for the following code: #include <limits.h> int x = -1; int main() { return INT_MIN % x; } However I can find no statement in the standard that this code invokes undefined or implementation-defined behavior. As far as I can tell, it's required to return 0. Is this a bug in gcc or am I missing some special exception the standard makes? Jens Gustedt You are probably right that this can be considered as a bug in the actual standard. The current draft addresses this problem: If the quotient a/b is representable, the expression (a/b)*b + a%b shall

Division ( / ) not giving my answer in postgresql

梦想与她 提交于 2019-12-03 07:21:06
问题 I have a table software and columns in it as dev_cost , sell_cost . If dev_cost is 16000 and sell_cost is 7500. How do I find the quantity of software to be sold in order to recover the dev_cost ? I have queried as below: select dev_cost / sell_cost from software ; It is returning 2 as the answer. But we need to get 3, right? What would be the query for that? Thanks in advance. 回答1: Your columns have integer types, and integer division truncates the result towards zero. To get an accurate

Non-restoring division algorithm

一曲冷凌霜 提交于 2019-12-03 06:53:12
Does anyone know the steps for dividing unsigned binary integers using non-restoring division? It's hard to find any good sources online. i.e if A = 101110 and B = 010111 how do we find A divided by B in non-restoring division? What do the registers look like in each step? Thanks! (My answer is a little late-reply. But I hope it will be useful for future visitors) Algorithm for Non-restoring division is given in below image : In this problem, Dividend (A) = 101110, ie 46, and Divisor (B) = 010111, ie 23. Initialization : Set Register A = Dividend = 000000 Set Register Q = Dividend = 101110 (

Why doesn't 'd /= d' throw a division by zero exception when d == 0?

霸气de小男生 提交于 2019-12-03 06:26:14
问题 I don't quite understand why I don't get a division by zero exception: int d = 0; d /= d; I expected to get a division by zero exception but instead d == 1 . Why doesn't d /= d throw a division by zero exception when d == 0 ? 回答1: C++ does not have a "Division by Zero" Exception to catch. The behavior you're observing is the result of Compiler optimizations: The compiler assumes Undefined Behavior doesn't happen Division by Zero in C++ is undefined behavior Therefore, code which can cause a

Evenly divide a dollar amount (decimal) by an integer

喜夏-厌秋 提交于 2019-12-03 04:25:38
问题 I need to write an accounting routine for a program I am building that will give me an even division of a decimal by an integer. So that for example: $143.13 / 5 = 28.62 28.62 28.63 28.63 28.63 I have seen the article here: Evenly divide in c#, but it seems like it only works for integer divisions. Any idea of an elegant solution to this problem? 回答1: Calculate the amounts one at a time, and subtract each amount from the total to make sure that you always have the correct total left: decimal

How to divide the value of pandas columns by the other column

旧城冷巷雨未停 提交于 2019-12-03 04:17:19
问题 I have a dataframe: >>> dt COL000 COL001 QT STK_ID RPT_Date STK000 20120331 2.6151 2.1467 1 20120630 4.0589 2.3442 2 20120930 4.4547 3.9204 3 20121231 4.1360 3.8559 4 STK001 20120331 -0.2178 0.9184 1 20120630 -1.9639 0.7900 2 20120930 -2.9147 1.0189 3 20121231 -2.5648 2.3743 4 STK002 20120331 -0.6426 0.9543 1 20120630 -0.3575 1.6085 2 20120930 -2.3549 0.7174 3 20121231 -3.4860 1.6324 4 And I want the columns values divided by 'QT' column, somewhat like this: dt = dt/dt.QT # pandas does not

Why is 0 divided by 0 an error?

て烟熏妆下的殇ゞ 提交于 2019-12-03 02:36:33
I have come across this problem in a calculation I do in my code, where the divisor is 0 if the divident is 0 too. In my code I return 0 for that case. I am wondering, while division by zero is generally undefined, why not make an exception for this case? My understanding why division by zero is undefined is basically that it cannot be reversed. However, I do not see this problem in the case 0/0. EDIT OK, so this question spawned a lot of discussion. I made the mistake of over-eagerly accepting an answer based on the fact that it received a lot of votes. I now accepted AakashM's answer ,

How to Calculate (a/b) %c where a,b and c are very large numbers [closed]

廉价感情. 提交于 2019-12-02 23:00:53
问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 5 years ago . I have a function f(x)=(1^1)*(2^2)*(3^3)*.....(x^x) i have to calculate (f(x)/(f(x-r)*f(r)))modulo c i can calculate f(x) and (f(x-r)*f(r)). assume f(x) is a and f(x-r)*f(r) is b. c is some number that is very larger. `` so i how can calculate (a/b)%c 回答1: your f(x) is just ᴨ (PI cumulative multiplication)