Invalid conversion 'void*' to 'struct*' [duplicate] - c

Question

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malloc - invalid conversion from void* to double*
(2 answers)
Closed 4 years ago.
I'm beginner in C. I'm trying to practice with solving some problems. And I'm getting this error when I compile my code.
[Error] invalid conversion from 'void*' to 'triangle*' [-fpermissive]
The code and purpose is explained below.
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
struct triangle
{
int a;
int b;
int c;
};
typedef struct triangle triangle;
//sort_by_area() function is here
int main()
{
int n;
scanf("%d", &n);
triangle *tr = malloc(n * sizeof(triangle));
for (int i = 0; i < n; i++) {
scanf("%d%d%d", &tr[i].a, &tr[i].b, &tr[i].c);
}
sort_by_area(tr, n);
for (int i = 0; i < n; i++) {
printf("%d %d %d\n", tr[i].a, tr[i].b, tr[i].c);
}
return 0;
}
As you can see I have structure and I tried to allocate memory for it with the quantity of input. And try to use it for sort_by_area function. But the problem is triangle *tr = malloc(n * sizeof(triangle)); line gives me the error mentioned above.
Also this code is working for online compilers. I tried to run this code on DEV C++ with default settings. I don't know about the versions and changing the versions of my compiler. I don't even know whether it is about the compiler version. But I am wondering why I'm getting this error. What is the logic behind.

Answer 1

This looks like C code, but you're compiling with a C++ compiler. As such, it complains on the line you mentioned because malloc returns a void * but you're assigning the result to a triangle *.
In C++ an explicit cast is required for this. In C, a void * may be implicitly converted to or from any object pointer type.
Since this appears to be C code and not C++ code, you should compile with a C compiler.

Answer 2

You compile this program as C++ program and this implicit conversion is not allowed in C++.
As I know dev C++ uses MinGW and you may use -xc option to compile program as C program or Settings -> language standard -> and choose the language standard needed

Answer 3

The code looks like C code but you are compiling it with C++ compiler.
Make sure that the file has proper extension for C++(not .c extension).
malloc() returns a (void *) pointer by default, so you have to explicitly cast the (void *) to (triangle *) in your code.
But if you are writing C++ code then I would recommend not to use malloc and free, instead try to use "new" operator in C++ since while instantiating objects it will call the constructors as well(unlike in malloc).
So to avoid complications go with new and delete in C++.
The code in C should look like (file a.c)
Compile using: gcc a.c -o a.o
Run using : ./a.o
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
struct triangle {
int a;
int b;
int c;
};
typedef struct triangle triangle;
int main() {
int n;
scanf("%d", &n);
triangle *tr = (triangle *)malloc(n * sizeof(triangle));
for (int i = 0; i < n; i++) {
scanf("%d%d%d", &tr[i].a, &tr[i].b, &tr[i].c);
}
//sort_by_area(tr, n);
for (int i = 0; i < n; i++) {
printf("%d %d %d\n", tr[i].a, tr[i].b, tr[i].c);
}
free(tr);
return 0;
}
The same code in C++ would look like (file a.cpp)
Compile using: g++ a.cpp -o a.o
Run using : ./a.o
#include <iostream>
using namespace std;
struct triangle {
int a;
int b;
int c;
};
int main() {
int n;
cin >> n;
triangle *tr = new triangle[n];
for (int i = 0; i < n; i++) {
cin >> tr[i].a >> tr[i].b >> tr[i].c;
}
// sort_by_area(tr, n);
for (int i = 0; i < n; i++) {
cout << tr[i].a << " " << tr[i].b << " " << tr[i].c << "\n";
}
delete [] tr;
return 0;
}

Answer 4

triangle *tr = (triangle*)malloc(n * sizeof(triangle));
Change the line as shown above. malloc returns generic pointer, so you need to explicitly typecast it to your desired pointer.
Refer this