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Why does C++ require a cast for malloc() but C doesn't?
In C language, the void pointers are converted implicitly to the object pointer type. The function malloc() returns void * in C89 standard. In earlier versions of C, malloc() returns char *. In C++ language, by default malloc() returns int value. So, the pointers are converted to object pointers using explicit casting.
The following is the syntax of allocating memory in C language.
pointer_name = malloc(size);
Here,
pointer_name − Any name given to the pointer.
size − Size of allocated memory in bytes.
The following is an example of malloc() in C language.
Example
#include <stdio.h>
#include <stdlib.h>
int main() {
int n = 4, i, *p, s = 0;
p = malloc(n * sizeof(int));
if(p == NULL) {
printf("\nError! memory not allocated.");
exit(0);
}
printf("\nEnter elements of array : ");
for(i = 0; i < n; ++i) {
scanf("%d", p + i);
s += *(p + i);
}
printf("\nSum : %d", s);
return 0;
}Output
Enter elements of array : 2 28 12 32 Sum : 74
In the above example in C language, if we do explicit casting, it will not show any error.
The following is the syntax of allocating memory in C++ language.
pointer_name = (cast-type*) malloc(size);
Here,
pointer_name − Any name given to the pointer.
cast-type − The datatype in which you want to cast the allocated memory by malloc().
size − Size of allocated memory in bytes.
The following is an example of malloc() in C++ language.
Example
#include <iostream>
using namespace std;
int main() {
int n = 4, i, *p, s = 0;
p = (int *)malloc(n * sizeof(int));
if(p == NULL) {
cout << "\nError! memory not allocated.";
exit(0);
}
cout << "\nEnter elements of array : ";
for(i = 0; i < n; ++i) {
cin >> (p + i);
s += *(p + i);
}
cout << "\nSum : ", s;
return 0;
}Output
Enter elements of array : 28 65 3 8 Sum : 104
In the above example in C++ language, if we will not do the explicit casting, the program will show the following error.
error: invalid conversion from ‘void*’ to ‘int*’ [-fpermissive] p = malloc(n * sizeof(int));
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