malloc-> how much memory is allocated? [duplicate] - c

Question

This question already has answers here:
I can use more memory than how much I've allocated with malloc(), why?
(17 answers)
Closed 9 years ago.
As I know that malloc allocate a specific number of bytes in the memory. However I am trying to use it and I allocate 4 bytes but it gives me NO error when I try to store more than 4 (up to 200 integers) elements in the array!! So in my code I don't need to use realloc!! I'm using Linux by the way. Finally I will pleased to hear any advice from you ... thanks in advance.
tmp.h :
#ifndef TMP_H
#define TMP_H
#define MAXLENGTH 4
#define GROWFACTOR 1.5
typedef struct stVector
{
int *vec;
int length;
int maxLength;
}Vector;
Vector newEmptyVector();
void addElement(Vector *vec, int elt);
#endif
tmp.c :
#include "stdio.h"
#include "stdlib.h"
#include "tmp.h"
Vector newEmptyVector()
{
Vector vec;
vec.vec = (int*) malloc(0);
printf("Allocating %d bytes\n", sizeof(int)*MAXLENGTH );
vec.length = 0;
vec.maxLength = MAXLENGTH;
return vec;
}
void addElement(Vector *vec, int elt)
{
/*if(vec->length == vec->maxLength)
{
vec->vec = (int*)realloc(vec->vec,sizeof(int)* vec->maxLength * GROWFACTOR);
vec->maxLength = vec->maxLength * GROWFACTOR;
}*/
vec->vec[vec->length++] = elt;
}
main.c :
#include"tmp.h"
int main(int argc, char const *argv[])
{
Vector vector = newEmptyVector();
printf("The length is %i and maxlength is ` `%i\n",vector.length,vector.maxLength);
addElement(&vector,5);
addElement(&vector,3);
addElement(&vector,1);
addElement(&vector,7);
printf("The length is %i and maxlength is ` `%i\n",vector.length,vector.maxLength);
addElement(&vector,51);
printf("The length is %i and maxlength is %i\n",vector.length,vector.maxLength);
for (int i = 0; i < 200; ++i)
{
addElement(&vector,i);
printf("The length is %i and maxlength is %i\n" ,vector.length, vector.maxLength);
}
return 0;
}

Answer 1

Using memory you haven't allocated invokes undefined behavior. Don't do that. In all likelyhood, Linux has give your program a page of memory, and you haven't overrun that yet. If you touch memory not allocated to your program the OS should cause your program to segfault. But it's possible that any other mallocing you do will also use parts of that page, and you'll end up corrupting your data.
Not having runtime checks for overrunning buffers is part of what makes C fast, but it puts more on the programmer not to do dumb things.

Answer 2

The fact that (simply because there is no bound checking in C) no error is raised does not mean that you can safely use memory outside requested bounds. You were lucky not to cause a segmentation fault, you have just fallen into a memory region that is not claimed by your malloc (let's say, it's not yours).
You can write there, but there is no guarantee that you won't be overwriting memory assigned to another malloc or, conversely, that the "extra" part will not be allocated to some other malloc. In your case, the memory region you are writing into appears not to be claimed (yet).

Answer 3

Regarding your specific issue:, I allocate 4 bytes but it gives me NO error when I try to store more than 4.
Keep in mind, something like:
int *anyVar = (int)malloc(0);
anyVar[0] = 12; //Will eventually invoke undefined behavior.
writing to memory you do not own will eventually invoke undefined behavior. The bad thing is that your results can seem good, and even repeatable for many runs of the code. But at some point, your code will fail.
This is how you should allocate: (by the way)
int numIntsInArray = 100;
int *anyVar = malloc(sizeof(int)*numIntsInArray);//note:do not cast output of malloc
anyVar[0] = 1;//first element of anyVar
anyVar[99] = 1000;//last element of anyVar
Do not forget to free all memory:
free(anyVar);
Other examples of undefined behavior in C & C++:
The examples of bad code below can be done, and you will likely get no compiler warnings, and may even get
expected results during run-time, but with this code, nothing is guaranteed. (exception: good examples)
char * p = "string"; // Badly formed C++11, deprecated C++98/C++03
p[0] = 'X'; // undefined behavior
Create an array instead:
char p[] = "string"; // Good
p[0] = 'X';
C++, you can create/use a standard string like this:
std::string s = "string"; // Good
s[0] = 'X';
Division by zero results in undefined behavior:
int x = 1;
return x / 0; // undefined behavior
Some pointer operations may lead to undefined behavior:
int arr[4] = {0, 1, 2, 3};
int* p = arr + 5; // undefined behavior
Leaving a non-void function without returning a value
int func(void)
{
//undefined behavior
}
Unspecified or implementation-defined behavior:
printf("%d %d\n", ++n, power(2, n)); //Bad
i = ++i + 1; //Bad
i = i + 1; // Okay