evaluation

C++ operator precedence in output stream

冷暖自知 提交于 2020-11-30 02:04:49
问题 int a = 1, b = 2; int c = a*b + b==0; // c = 0 cout << a*b + b==0; // outputs 4 c evaluates to 0 because the operator precedence of the * and + operators is higher than == as a result of which c essentially evaluates to (a*b+b)==0 which is false. Why does putting the same expression in a cout statement output 4? 回答1: Because the precedence of these operators are operator* > operator+ > operator<< > operator== . Then cout << a*b + b==0; is equivalent with (cout << ((a*b) + b)) == 0; . Then the

C++ operator precedence in output stream

眉间皱痕 提交于 2020-11-30 02:02:21
问题 int a = 1, b = 2; int c = a*b + b==0; // c = 0 cout << a*b + b==0; // outputs 4 c evaluates to 0 because the operator precedence of the * and + operators is higher than == as a result of which c essentially evaluates to (a*b+b)==0 which is false. Why does putting the same expression in a cout statement output 4? 回答1: Because the precedence of these operators are operator* > operator+ > operator<< > operator== . Then cout << a*b + b==0; is equivalent with (cout << ((a*b) + b)) == 0; . Then the

C++ operator precedence in output stream

落花浮王杯 提交于 2020-11-30 02:00:28
问题 int a = 1, b = 2; int c = a*b + b==0; // c = 0 cout << a*b + b==0; // outputs 4 c evaluates to 0 because the operator precedence of the * and + operators is higher than == as a result of which c essentially evaluates to (a*b+b)==0 which is false. Why does putting the same expression in a cout statement output 4? 回答1: Because the precedence of these operators are operator* > operator+ > operator<< > operator== . Then cout << a*b + b==0; is equivalent with (cout << ((a*b) + b)) == 0; . Then the

eval(string) to C# code

只愿长相守 提交于 2020-07-03 07:04:08
问题 Is it possible to evaluate the following in C# at runtime I have a class that contains 3 properties ( Field , Operator , Value ) rule.Field; rule.Operator; rule.Value; this is my rule class... Now I have a loop foreach(item in items) { // here I want to create a dynamic expression to evaluate at runtime // something like if (item.[rule.field] [rule.operator] [rule.value]) { do work } } I just don't know the syntax, or if its possible in C#, I know in JS its possible but that's not a compiled

How to evaluate functions in GDB without painful typecasting?

北战南征 提交于 2020-06-01 03:40:13
问题 In an answer to How to evaluate functions in GDB? I found the recipe to call a function outside of my program called floor from within GDB like this: (gdb) p floor $20 = {<text variable, no debug info>} 0x38e10197b0 <floor> (gdb) p ((double(*)(double))floor)(2.9999) $21 = 2 (gdb) p ((double(*)(double))floor)(2.000001) $22 = 2 (gdb) What do I need to do, short of a compiler upgrade which is not an option for me at the moment, in order for me to call the function in the same manner in which I

How to evaluate functions in GDB without painful typecasting?

孤者浪人 提交于 2020-06-01 03:40:04
问题 In an answer to How to evaluate functions in GDB? I found the recipe to call a function outside of my program called floor from within GDB like this: (gdb) p floor $20 = {<text variable, no debug info>} 0x38e10197b0 <floor> (gdb) p ((double(*)(double))floor)(2.9999) $21 = 2 (gdb) p ((double(*)(double))floor)(2.000001) $22 = 2 (gdb) What do I need to do, short of a compiler upgrade which is not an option for me at the moment, in order for me to call the function in the same manner in which I

How to evaluate functions in GDB without painful typecasting?

别等时光非礼了梦想. 提交于 2020-06-01 03:39:59
问题 In an answer to How to evaluate functions in GDB? I found the recipe to call a function outside of my program called floor from within GDB like this: (gdb) p floor $20 = {<text variable, no debug info>} 0x38e10197b0 <floor> (gdb) p ((double(*)(double))floor)(2.9999) $21 = 2 (gdb) p ((double(*)(double))floor)(2.000001) $22 = 2 (gdb) What do I need to do, short of a compiler upgrade which is not an option for me at the moment, in order for me to call the function in the same manner in which I

Evaluating a symbolic function

半世苍凉 提交于 2020-03-15 09:37:29
问题 I want to find cos(5) . Why is this expression invalid: syms x f=sin(x) disp(diff(f)(5)) The error is Line: 3 Column: 12 Indexing with parentheses '()' must appear as the last operation of a valid indexing expression. 回答1: Your error has nothing to do with symbolic variables. It is caused by the statement diff(f)(5) - which is not something MATLAB syntax allows (as of R2019b). MATLAB interprets this as the user trying to access the 5th element of some intermediate result. If you want to know

Threshold for evaluation deep learning

两盒软妹~` 提交于 2020-03-05 05:17:27
问题 I would like if possible how to make this Threshold for Evaluation and validation of created R-CNN object Detector, i tried to make it in the attached scripts but it does not work, I want to make Threshold for score that like below 0.58 that score and bboxes should not be appeared Herein the code:- load('gTruth.mat') output = selectLabels(gTruth,'signal'); if ~isfolder(fullfile('EvaluationData')) mkdir EvaluationData addpath('EvaluationData'); evaluationData = objectDetectorTrainingData