To repeat the answers you had in the comments already, the difference is the type returned void foo() { … } Is a function that returns nothing, while void *bar() { … } void* bar() { … } // Identical Return a void pointer. Swapping the space position before or after the * makes no … Read more
The problem is with the pointer assignments as you build the slice you are trying to reference. The addresses keep changing. func main(){ var lump []int // A loop to build a slice of `int`’s from 0 size to 8 size // and print each index address for i:= 0; i < 8; i++{ lump … Read more
Your program is causing memory leak. You should use an array of ADI*, not an array of ADI, to store what is returned from adjancencyList(). Using subscripting looks better than incrementing in this case. One more tips is that A = &( A[0] ); does virtually nothing. Also note that they say you shouldn’t cast … Read more
Error was in v_push_back implementation : v->e_array = check_a(memcpy((void*)((char*)v->e_array + (v->no_e * v->e_sz)) ,new_val, v->e_sz)); //pointer has not to be saved check_a(memcpy((void*)((char*)v->e_array + (v->no_e * v->e_sz)) ,new_val, v->e_sz)); Because memcpy returns a pointer , in this case, to the last element of the array so we would have lost access to all previous memory solved … Read more
++ has a higher precedence (meaning it binds to p tighter) than * therefore *p++ is equivalent to *(p++). Something similar to this is the difference between *p[] and (*p)[]. [] has a higher precedence than * therefore *p[] is equivalent to *(p[]) which makes a list a pointers but something like (*p)[] explicitly says … Read more
You probably want this: #include <stdio.h> #include <string.h> int main(void) { char temp[256]; char *st[3]; // array of three pointers to char for (int i = 0; i < 3; i++) { scanf(“%255s”, temp); // prevents potential buffer overflow st[i] = strdup(temp); } for(int i = 0; i < 3; i++) { printf(“%s\n”, st[i]); free(st[i]); … Read more
At these two lines: *min=t[0]; *max=t[0]; Your code assumes that t[0] is an even number. However, that’s not necessarily true. One fix would be: *min = INT_MIN; *max = INT_MAX; 1 solved Why is this program going this way? C
What does this function do? It concatenates two strings, i.e. it’s doing the same as the standard strcat function. See https://man7.org/linux/man-pages/man3/strcat.3.html Assume the input is “Hello World”. Just when the function is called, some where in memory it looks like: String1: Hello\0 ^ | s1 String2: World\0 ^ | s2 Now this part while (*s1 … Read more
C11, 6.3.2.3, paragraphs 5 and 6: An integer may be converted to any pointer type. Except as previously specified, the result is implementation-defined, might not be correctly aligned, might not point to an entity of the referenced type, and might be a trap representation. Any pointer type may be converted to an integer type. Except … Read more
No, it’s a syntax error. It can’t be part of a multiplication since there cannot be an identifier with the name of a keyword. The only way to make this a valid part of a C program is as part of a string literal as part of a multibyte character constant as part of a … Read more
Because when you pass an array to a function, it decays to a pointer, more precisely to a pointer to the first element. In your example, the argument arr is actually a pointer and not an array (which is also why you can’t use sizeof on the “array”, that gets the size of the pointer … Read more
When is using the “this pointer” useful in a C++ program When you need a pointer to the current object. One of the most often usages of this that I have seen is to make self assignment a noop. Foo& Foo::operator=(Foo const& rhs) { if ( this != &rhs ) { // Assign only when … Read more
Your example is basically equivalent to this: void changeString(string **strPtr) { strPtr = new string*[1]; *strPtr = new string(“hello”); //cout << strPtr[0][0]; } int main() { string *strPtr; changeString(&strPtr); //cout << strPtr[0]; return 0; } And that is basically equivalent to this: void changeString(string *strPtr) { strPtr = new string(“hello”); // Changes the pointer, not … Read more
First line: You create a pointer variable to a char, then you initialize it to address zero (NULL pointer). Second line: you try to write a zero to the address where pointer is pointing to. Address zero is outside of your process’ writable virtual memory area, so you get a segmentation fault. solved Pointer making … Read more
circleType::setRadius; double radius = circle.getRadius; The first line doesn’t do anything. The second line tries to set a variable of type double equal to a function. Perhaps you want to call these functions? This is what non-operator member function calls look like in C++: double area = circle.areaCir(radius); double circumferance = circle.circumCir(radius); circle.printCir(circumferance, area); So … Read more