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If I allocate an array of some primitive type e.g.
double *v = new double[10];
I need to know, what the inital value of the array entries will be.
Is it specified in the standard or compiler dependend and where can I find it.
Thanks, Johannes
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The “malloc” or “memory allocation” method in C is used to dynamically allocate a single large block of memory with the specified size. It returns a pointer of type void which can be cast into a pointer of any form.
C uses the malloc() and calloc() function to allocate memory dynamically at run time and uses a free() function to free dynamically allocated memory. C++ supports these functions and also has two operators new and delete, that perform the task of allocating and freeing the memory in a better and easier way.
C Program : // C Program to dynamically allocate an int ptr #include <stdio.h> #include <stdlib.h> int main() { // Dynamically allocated variable, sizeof(char) = 1 byte. char *ptr = (char *)malloc(sizeof(char)); if (ptr == NULL) { printf("Memory Error!\n"); } else { *ptr = 'S'; printf("%c", *ptr); } return 0; }
No, the array contents are not initialized. You need to use double *v = new double[10](); to have the default value of 0 for each element (Notice ()).
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