You didn’t copy the program correctly. This is the real program that reproduces the bug: #include <stdio.h> #define TRUE 1 #define FALSE 0 int function_returning_false() {return FALSE;} int main() { if (function_returning_false) { // note, no () printf(“function returned true\n”); } } So, it’s not calling the function – it is passing the function as … Read more
I don’t know where you get your analysis from. Most parts like the abstract syntax tree is kept because it is used in all different passes. It might be that some, especially simple compilers don’t free stuff because it’s not considered necessary for a C compiler. It’s a one shot compilation unit operation and than … Read more
The actual type is hidden as it’s local inside the function, so you can’t explicitly use it. You should however be able to use decltype as in decltype(create<int>()) v = create<int>(); I fail to see a reason to do like this though, when auto works. 1 solved How to specify type of a constexpr function … Read more
Headers included with one compiler are typically tailored to that compiler implementation and will not necessarily work correctly with a different compiler. So generally speaking you won’t be able use the headers that come with Visual Studio with another compiler. 7 solved Why won’t clang compile this source code that works in VS2012?
Actually, the function returns a value. Maybe you want to see it, so just print the result: #include <stdio.h> #include <cs50.h> int calcrectarea(int len,int wid){ return len*wid; } int main(){ printf(“enter length here:\n”); int x; scanf(“%d”,&x); printf(“enter width here:\n”); int y; scanf(“%d”, &y); printf(“area is: %d”, calcrectarea(x,y)); } solved The function is not returning in … Read more
Forgetting the return value results in the control flow reaching the end of a function. The C and C++ standards both describe this case. Note that main is an exception to this case and is described separately. Flowing off the end of a function is equivalent to a return with no value; this results in … Read more
You assigned outputMessage, which is an array and is converted to a pointer to the first element of the array, to messagePtr, so messagePtr no longer points at what is allcated via malloc() or its family. Passing what is not NULL and is not allocated via memory management functions such as malloc() invokes undefined behavior. … Read more