ptr is the actual pointer, while *ptr is whatever it is pointing at, so *ptr=&var does not really make any sense, unless it’s a pointer to a pointer. It’s either ptr=&var or *ptr=var If you really want to assign a variable to a pointer, it is possible with casting. This compiles, but I cannot see … Read more
When you declare char string[]=”Hello” you declare an array of characters. string points to the first element of the array. * is known as the dereferencing operator. Suppose ptr is an integer pointer. *ptr will give you the content of the memory location pointed by the pointer ptr. At the time of declaration you have … Read more
Yes they are exactly the same. The first code in words: SomeClass pointer sc = new SomeClass(); return SomeClass pointer sc The second code in words SomeClass pointer sc = new SomeClass(); SomeClass pointer rSc = SomeClass pointer sc return SomeClass pointer rSc As you can read/ see for yourself there is no difference since … Read more
* is both a binary and a unary operator in C and it means different things in different context. Based on the code you have provided: *leaf = (struct node*) malloc( sizeof( struct node ) ); Here the void * (void pointer) that malloc returns is being casted to a pointer to struct node, I … Read more
Is this Correct? No. Mixing new and malloc() is not a good idea. If you must use dynamic memory allocation, then read this: How do I declare a 2d array in C++ using new? Otherwise, use an std::vector. 12 solved c++ : An array of the pointer float type with dynamic size
The problem is in foreach templated function: first parameter can be a pointer to the array or it can just be static array array with undefined size second parameter can be a function pointer or a reference to the function (which should accept constant element of same type as array elements by reference, as it … Read more
newarray is an automatic local variable. It will no longer exists once function return. Returning pointer to an automatic local variable invoke undefined behaviour and in this case nothing good can be expected. You can allocate memory diynamically and then return pointer to it int *newArray = malloc(sizeof(int)*size); 2 solved Returning pointer to array doesn’t … Read more
Based on the threads Ive read on here, setting a pointer to deadbeef will make the pointer null. No, that is incorrect. Although deadbeef is a common “nonsense pattern”, it is definitely not equal null. A common reason to set pointers (more generally, a memory area) to a known and unusual pattern, such as deadbeef, … Read more
Okay to explain it in a more complete way. The most other guys had already wrote that you have to use ‘->’ when ever you have a pointer. But you could also do this with ‘.’, to do this you must respect the priority of the operators. We need the ‘*’ to get the content … Read more
“blah” is a const char [5]. In the first line, that array is decayed into a pointer to be stored in your variable as a pointer to the first element. It is also a pointer to non-const characters that points to const characters. It should be: const char *sz1 = “blah”; In the second (thanks … Read more
*x = 1; means that x contains a memory address, assign 1 to that memory address. x = 1; means assign 1 to the variable x. solved What is the difference between “*x = 1” and “x = 1” in C?
Let’s use std::vector: #include <iostream> #include <vector> #include <iomanip> typedef std::vector<double> DoubleArray; DoubleArray v_scalar_prod(double a, double *b, int n) { DoubleArray res(n); for (int i = 0; i < n; i++) { res[i] = a*b[i]; std::cout << std::setprecision(10) << “res = ” << res[i] << ‘\n’; } return res; } int main() { double y[3] … Read more
You are passing the arrays to sameNumberCheckOfTwoArrays as pointers. In this process, the “array-ness” is dropped. Your function only sees an int* and reports the size accordingly. 3 solved Something went wrong with passing pointer to a function
Three problems here. First scanf(ifp,”%lf %lf\n”,&theta,&Cla); You’re calling scanf when you look like you want to use fscanf: fscanf(ifp,”%lf %lf\n”,&theta,&Cla); Second is the file you’re opening: ifp=fopen(argv[0],”r”); argv[0] is the name of the running executable, which is probably not what you want. If you want the first argument passed in, use argv[1] ifp=fopen(argv[1],”r”); Last is … Read more
Point 1: Nested functions are not standard C. They are supported as GCC extension.. Point 2: printf(“&b is:%s\n”,&b); is wrong and invokes UB, because of improper format specifier. You need to change that to printf(“&b is:%p\n”,(void *)&b); Point 3: &(&a) is wrong. the operand for & needs to be an lvalue, not another address, which … Read more