operator-precedence

问题 I have this statement: return *local_stack_var2++ + 42; Would these be the proper steps when breaking it down: 1. Dereference local_stack_var2 2. Add 42 to the dereferenced local_stack_var2 (function will actually return this value) 3. Before the function is over, it will activate the post-increment, incrementing the value of the object pointed to by local_stack_var2 So in code format, it would look kind of something like this? int temp = *local_stack_var2 //step 1; int returnValue = temp +

问题 This question already has answers here : Calculation error with pow operator (4 answers) Closed 6 years ago . In Python >>> i = 3 >>> -i**4 -81 Why is -i**4 not evaluated as (-i)**4 , but as -(i**4) ? I suppose one could argue that raising to a power takes precedence over (implicit) multiplication of i with minus one (i.e. you should read -1*i**4 ). But where I learned math, -i**n with n even and i positive, should come out positive. 回答1: The ** operator binds more tightly than the - operator

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

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

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

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

来源： https://stackoverflow.com/questions/63670099/why-this-loop-runs-10-times-instead-of-5

来源： https://stackoverflow.com/questions/63670099/why-this-loop-runs-10-times-instead-of-5

来源： https://stackoverflow.com/questions/22289093/can-you-impose-object-precedence-for-overloaded-operators-in-python

来源： https://stackoverflow.com/questions/49895131/logical-vs-assignment-operator-precedence-in-php