Problem in malloc

Quote:

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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?

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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:

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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:

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int *integers;

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the same as

Code:

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int *integers[i]?

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where i > 0?

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

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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:

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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.

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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:

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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?

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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:

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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.

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it has the same output as if he did:

Code:

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

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Quote:

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Originally Posted by abraham2119

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

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

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

The OP delcared

Code:

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int *integers;

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which is the same as an array declaration

Code:

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int integers[0];

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But when he used

Code:

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integers[i]

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

Quote:

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Originally Posted by happyclown

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

The OP delcared

Code:

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int *integers;

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which is the same as an array declaration

Code:

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int integers[0];

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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.