Visual Studio 2015 doesn’t support the C++17 standard. Prior to C++17 standard you can’t have the std::shared_ptr<T[]> pointer. But even in C++17 the std::make_shared function doesn’t support array types so you would have to use the boost::make_shared instead. Another alternative is to use the unique pointer in combination with the std::make_unique which does support the … Read more
Just take int a =10; int *p = &a; a is a variable which holds value 10. The address in which this value is stored is given by &a. Now we have a pointer p , basically pointer points to some memory location and in this case it is pointing to memory location &a . … Read more
In foo() and foo2() you are returning pointers to local variables. These locals have automatic storage and it is undefined behavior to use a pointer to them once they go out of scope. Therefor p and p2 contain nothing useful. It might work, it might not. I can’t understand your question very well, so I … Read more
Does this mean that x holds the address of first element of the string,i.e, the character ‘g’ ? Yes. is it the placeholder %s that’s instructing the printf to print the string literals ? Yes. To be more specific, %s is not limited for string literals. It is for printing null terminated srings – which … Read more
From Wikipedia: Short: Short signed integer type. Capable of containing at least the [−32,767, +32,767] range;[3][4] thus, it is at least 16 bits in size. The negative value is −32767 (not −32768) due to the one’s-complement and sign-magnitude representations allowed by the standard, though the two’s-complement representation is much more common. Double: Real floating-point type, … Read more
Let’s look at the first two lines of your code: str = (char *) malloc(15); strcpy(str, “63tczfqV4nqB2YnH9iFJbGvGyyDkvK341rrj0G0mo1PEYniOVHejVIFIQnJzHSSMRbuyGMCZ4M5HFMV4y1q4QgYqyxp2XkTjxaolKTkaw1r25S2Emz061tw1”); At this point, you have broken the rules of the C language. strcpy will write past the end of str which causes undefined behavior. Everything that happens after this point is kinda up in the air. 5 solved … Read more
The three are same. The only differences are, The position of the asterisk and the whitespace around them(which is done according to one’s preference) The string literals “blah” could be stored in the same memory location or different locations. From the C11 standard, 6.4.5 String literals […] It is unspecified whether these arrays are distinct … Read more
myst1r is an array of char that holds a copy of its initializer. mystr is a pointer to char that points to its initializer. In both cases, the initializer is a literal string. The difference is that you can modify the contents of an array, but you cannot modify the contents of a literal string. … Read more
#define int int* will replace int *p, q as int* *p, q. So Here p is double pointer to int and q is of type int. For example consider the below program of your same logic in char #include<stdio.h> #define char char* main() { char *p,q; printf(“%d, %d\n”, sizeof(p), sizeof(q)); } Output is 4, 1 … Read more
This syntax in invalid. You cannot simply write code inside a class definition. Instead, you can write a function and your code will work inside it: class Foo { public: void bar() { int * p = new int; *p = 5; } }; int main(int argc, char ** argv) { Foo f; f.bar(); } … Read more
No! You’re talking some mix-up of C and C++. This is C code. C does not have references. The & operator can be applied to objects in order to retrieve their address in memory, yielding a pointer, in both C and C++. It ain’t got nothing to do with references. Even in C++. In C++, … Read more
This is a popular question for people who learn python. The answer is that python has no pointer but using id(). And id() return C language memory address in CPython interpreter. Everything in python is a python object and id() is the identity of the python object. When you write python code you need to … Read more
The 3 functions are the same. The difference is the argument you pass to them. &p: get the reference (memory adress) of the variable p. *p: assumes that p is a memory address and get the value that is stored in that position. In this case, you will get a sigsev because you are trying … Read more
Java’s pointers (so called by the language specification, while I believe Java programmers mostly know them as “references”) do not support pointer arithmetic. And Java has no way to do pointer arithmetic. If you want to do such low-level stuff in Java, one way is to use JNI (the Java Native Interface) to access code … Read more
This char s[]=”TvNnFs”,*p; where s is an array of characters and p is character pointer, is looks like below s[0] s[1] s[2] s[3] s[4] s[5] s[6] —————————————— | T | v | N | n | F | s | \0 | —————————————— s 0x100 0x101 0x102 0x103 0x104 0x105 0x106.. (assume 0x100 is base … Read more