Problem in malloc

was studying about dynamic memory today but when i did this code it didnt compile with
msv6 but compiled with dev C++
anyways here the code

Code:

---

#include <stdio.h>
#include <stdlib.h>
int main()
{
int *integers;
int required=3;
int i;
integers=malloc(required*sizeof(int));
if(integers==NULL)
{
puts("Malloc failed");
exit(1);
}
for(i=0;i<required;i++)
{
printf("Please enter number #%d: ",i+1);
getchar();
scanf("%d",&integers[i]);
}
for(i=0;i<required;i++)
{
        printf("Number #%d was %d\n",i+1,integers[i]);
        getchar();
}
return 0;
}

---

What errors did you get?

For those of us that don't have either Dev C++, or msV6, or who might not even be on Windows, pray tell, what's the error you get?

D:\c_LESSONS_TUTRIALS_MADE_BY_ME\malloc\main.cpp(8 ) : error C2440: '=' : cannot convert from 'void *' to 'int *'

yaah srry to forgot to post error

you are compiling your C code as C++ (C supports implicit casts from void pointers, C++ doesn't).

I think you need an array of pointesr to do the job:

Code:

---

int *integers[3];

---

You have only allocated memory to integers, which is really integers[0], therefore integers[1] & integers[2] have not had memory allocated.

================

EDIT: If

Code:

---

int *integers;

---

is declared, but

Code:

---

integers[i]

---

is used in the program, shouldn't the compiler give a warning?

===================

So you have to use malloc to allocate memory to all 3 pointers, right before the user inputs each number:

Code:

---

for ( i = 0; i < required; i++ )
{
    printf("Please enter number #%d: ",i+1);
  intergers[i] = malloc( sizeof(int) );
  scanf("%d",&integers[i]);
}

---

and then you have to free the memory malloc has allocated to each pointer:

Code:

---

for(i=0;i<required;i++)
{
        printf("Number #%d was %d\n",i+1,integers[i]);
        getchar();
        free(integers[i]);
}

---

:)

Quote:

---

Originally Posted by cyberfish

you are compiling your C code as C++ (C supports implicit casts from void pointers, C++ doesn't).

---

And to fix that, all you actually need to do is rename the file from something.cpp to something.c and MS VC will automatically use the C flavour of it's compiler, rather than the C++ flavour.

--
Mats

yah thanks it compiles cool now thanks mats i dunno why in msv6 you cant compile as .c alawys .cpp

Quote:

---

Originally Posted by lolguy

yah thanks it compiles cool now thanks mats i dunno why in msv6 you cant compile as .c alawys .cpp

---

You CAN compile C as well as C++ in MSVC6, you just have to make sure that when you create the file, you name it as .c rather than .cpp which the IDE insists on if you don't give it a different extension.

--
Mats

yah i have to rename but when i start new project and new file there option only C++ source file there is no C source file

Quote:

---

Originally Posted by lolguy

yah i have to rename but when i start new project and new file there option only C++ source file there is no C source file

---

Yeah. I use the "Add new item..." in the Project view instead of the New File ... dialog - but I'm also using the next release of Visual Studio (.Net) rather than 6 - I think I still have 6 somewhere on my machine, but I'm not sure. ... Actually, I have VS 5.0 - and I think VS 6 is similar - when you select "New file ...", just click on the C++ file type, and then make the extension .c, and you should be fine.

--
Mats

oh cool thanks alot mate

Quote:

---

Originally Posted by happyclown

================

EDIT: If

Code:

---

int *integers;

---

is declared, but

Code:

---

integers[i]

---

is used in the program, shouldn't the compiler give a warning?

---

Well, but, why? Given that the OP's usage was perfectly correct, and in fact declaring a pointer and then mallocing memory is pretty much the only way to do dynamic memory allocation in C, then a whole heck of a lot of code would fall under this restriction.

Well, is it to okay to declare a pointer of type int( int *integers), and then use an array of type int[integers[i]), which hasn't been declared? But the compiler didn't give a warning about the undeclared array?

EDIT: so is

Code:

---

int *integers;

---

the same as

Code:

---

int integers[i]?

---

where i > 0?

I am a newbie, so maybe I am missing something(s)? :)

Quote:

---

Originally Posted by happyclown

Well, is it to okay to declare a pointer of type int( int *integers), and then use an array of type int[integers[i]), which hasn't been declared? But the compiler didn't give a warning about the undeclared array?

---

you mean, which hasn't been initialized?
if so, then the compiler will not give you an error during compile time, but it probably will during runtime.

EDIT:

Quote:

---

Originally Posted by happyclown

Well, is it to okay to declare a pointer of type int( int *integers), and then use an array of type int[integers[i]), which hasn't been declared? But the compiler didn't give a warning about the undeclared array?

EDIT: so is

Code:

---

int *integers;

---

the same as

Code:

---

int *integers[i]?

---

where i > 0?

I am a newbie, so maybe I am missing something(s)? :)

---

no.
int *integers is similar but NOT the same as int integers[].

one is a pointer of integers while the other is an array of integers.

btw, if i am not answering your question it's because i'm not understanding it correctly. maybe you should rephrase what you said:
because the question i understood you asked made no sense at all since you're asking if:
int *integers is the same as int *integers[0] which is completely wrong syntax.

Quote:

---

Originally Posted by abraham2119

you mean, which hasn't been initialized?
if so, then the compiler will not give you an error during compile time, but it probably will during runtime.

---

I mean the op declared a single pointer of type int, but then uses an undeclared array of an unknown type. :)

EDIT: but the compiler didn't give a warning. Should a warning have been given?

Quote:

---

Originally Posted by happyclown

I mean the op declared a single pointer of type int, but then uses an undeclared array of an unknown type. :)

EDIT: but the compiler didn't give a warning. Should a warning have been given?

---

ohhh i understand your question now. the op was calling the address of a component in his integers pointer. pointers are very similar but NOT the same to arrays.

when he does:

Code:

---

int *integers; //int pointer declaration
integers = malloc(10); //allocating memory (10 components) to the address which the pointer is pointing to
integers[0] = 25; //assigning a component in the index 0 of the integers pointer.
printf("%d", integers[0]); //printing component in index 0 of the 'integers' pointer.

---

it has the same output as if he did:

Code:

---

int integers[10];
integers[0] = 25;
printf("%d", integers[0]);

---

Quote:

---

Originally Posted by abraham2119

int *integers is the same as int *integers[0] which is completely wrong syntax.

---

Yep, my mistake. I've edited my previous post.

Ok, here's a rephrase of all my verbage above.

The OP delcared

Code:

---

int *integers;

---

which is the same as an array declaration

Code:

---

int integers[0];

---

But when he used

Code:

---

integers[i]

---

where i > 0, the array was out of bounds, but the compiler didn't give a warning. :)

Quote:

---

Originally Posted by happyclown

Ok, here's a rephrase of all my verbage above.

The OP delcared

Code:

---

int *integers;

---

which is the same as an array declaration

Code:

---

int integers[0];

---

---

You may claim that, but that doesn't mean it has any basis in fact (since it doesn't). Declaring int *integers does nothing more nor less than say that the variable "integers" is of type "int *", meaning a pointer to int.