You are invoking some very undefined behavior. The variable a in the function is at an address (probably on the stack) that is normally only accessible to the function. Decrementing that address results in an undefined location. You don’t know what’s there at all, so you have no idea what incrementing it by 8 will … Read more
Introduction If you are a C programmer, you may have encountered a situation where you wrote a code and got a wrong answer. This can be very frustrating and can be difficult to debug. In this article, we will discuss some of the common reasons why you may get a wrong answer in your C … Read more
List test = NULL; one(&test); test is a pointer to struct node. We pass the address of that pointer to one, allowing one to change test (letting it return a new pointer is a cleaner way to do this). one then executes this: two(&self); two is now being passed the address of the address of … Read more
with your “warning” you also get a string number, where “warning” appears. please give the string where an error appears I did not see at first time your warnings just replace the -1 to NULL or replace -1 to (void *)-1 course standard should be written to NULL, but if you really want, you can … Read more
I’m not sure what your question is, but this code probably doesn’t do what you want: printf(“main Pointer to struct.” EOL); for (int i=0; i<count; i++) { printf(“index: %d useless1: %d” EOL, i, b->useless1); printf(“index: %d useless2: %d” EOL, i, b->useless2); b++; } printf(EOL); printf(“main Index into array of structures.” EOL); for (int i=0; i<count; … Read more
What does the arguments (char * &str1, char * &str2) specify in the swap function? they specify that parameters are of type reference to pointer to char. This means that function might change the value of the variable that was provided as argument to this function. What if they are replaced by (char &str1, char … Read more
You are returning a pointer to the end of the string, and not the beginning of the string. You need to return the pointer that malloc gave you. That’s what you placed in new_str in your initial assignment to new_str. But instead of returning that, you modify that pointer and then return it. There are … Read more
It seems that you want to change the database type, from: std::vector <Employee> *EmployeeDB; this is a pointer to a vector, holding Employee instances. This means that the vector contains copies of employees, probably not what you want. To: std::vector <Employee*> EmployeeDB; This is a vector instance, holding pointers to Employees. You also should take … Read more
OP tagged C++ so here goes.. std::stringstream ss; ss << f1; std::string s = ss.str(); ptr = s.c_str(); 3 solved How to convert float to string using single char pointer? [closed]
The starting asterisk indicates that the data is to be read from the stream but ignored (i.e. it is not stored in the location pointed by an argument). Reference: http://www.cplusplus.com/reference/cstdio/scanf/ So the data would not be saved in the variable. 1 solved What does %*c do when getting input from stdin? [closed]
The obvious way: Picture temp; temp = Picture1; Picture1 = Picture2; Picture2 = temp; solved Re-referencing objects
This seams to be bad quality code! Maybe not dangerous, as const appears in prototype. myFunc(const void * p) accepts a pointer to anything and const should mean it won’t touch it. Now, st is a pointer to myStruct, so st->arr is the value of arr member and &st->arr is memory address of arr member. … Read more
ZeroA and so one are pointers to an array of char elements. &ZeroA holds the address of the pointer to the array ZeroA and so to hold it you need char ** The correct way to do it in your example is like this: char *XYZ[11]={ZeroA,OneA,TwoA,ThreeA,FourA,FiveA,SixA,SevenA,EightA,NineA,TenA}; 15 solved Warning in array of pointers “initialization from … Read more
This is just a manifestation of data locality. It takes less time to look at something at the next page of a book than something at the 800th next page. Try for yourself at home. 2 solved Dereferencing iterators performance
Returning a pointer isn’t a problem as long as it points to memory that is still in scope, such as dynamically allocated memory, global data, or local variables that still exist further up the call stack. The problem with the above code is that you’re dereferencing a pointer without assigning anything to it. This: *ret_val … Read more