#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void foo() {
char *a_heap_pointer;
a_heap_pointer = (char*)malloc(1 * sizeof(char));
strcpy(a_heap_pointer, "a");
printf("%s\n", a_heap_pointer);
while (1){
++a_heap_pointer;
a_heap_pointer = (char*)malloc(1 * sizeof(char));
strcpy(a_heap_pointer, "b");
printf("%s\n", a_heap_pointer);
}
}
int main (void) {
foo();
return 0;
}
Hi there, I'm trying to make my C program cause a segmentation fault by incrementing a pointer that points at the heap, by 1 byte, repeatedly, until the program crashes. However, as my code stands now it just runs indefinitely? I presume because the heap is so large that it would take awhile for it to run out. How can I make my program break the heap?
Printing on the screen is very slow compared to in-memory operations, so if you remove the printfs, you will probably not have to wait as long.
But modern computers have a lot of memory, and when you take into account swap space on disk it can be many hundreds of gigabytes. Prepare to wait for a long time.
Also, in C, strings are NUL-terminated, so the strings "a" and "b" both require two chars storage. Therefore your calls to strcpy overwrite memory that may be used for other things, so your program's behaviour is undefined. Anything can happen, for example that it gets stuck and never terminates.
And a side note: I'm not sure what effect you expect ++a_heap_pointer; to have? You increment that pointer, but then you immediately replace the incremented value by the return value from malloc, so the result of the increment operation will never be used.
Related
I'm running these code
#include <stdio.h>
void Crash(char * cData){
cData[2] = 100;
}
int main() {
char cData[2] = {1,2};
Crash(&cData[0]);
printf("%d\n",cData[1]);
return 0;
}
I expected the program to crash since cData[2] = 100;(of Crash()) will change the return address of Crash function. I believe that the memory position right next to cData[1](of main()) keeps the return address of Crash function. So after the Crash function finished execution, it will take the value in the return address(which is 100 now) and continue to execute other code. So shouldn't doing so suppose to cause the program to crash?
Your program has undefined behavior, which can be anything, including no crash at all and even expected behavior.
In your particular case, there is a chance the array char cData[2] = {1,2}; occupies space on the stack that is padded with 2 extra bytes before other important pieces of information such as the return address or the saved stack frame pointer. Modifying one of these bytes would have no noticeable effect. Try modifying cData[4] or cData[8], etc. but don't blame me for undesirable side effects.
so I'm quite new in this, sorry if it sound like a dumb question
I'm trying to understand malloc, and create a very simple program which will print "ABC" using ASCII code
here is my code (what our professor taught us) so far
char *i;
i = malloc(sizeof(char)*4);
*i = 65;
*(i+1) = 66;
*(i+2) = 67;
*(i+3) = '\0';
what I don't understand is, why do I have to put malloc there?
the professor told us the program won't run without the malloc,
but when I tried and run it without the malloc, the program run just fine.
so what's the function of malloc there?
am I even using it right?
any help and or explanation would be really appreciated
the professor told us the program won't run without the malloc
This is not quite true, the correct wording would be: "The program's behavior is undefined without malloc()".
The reason for this is that
char *i;
just declares a pointer to a char, but there's no initialization -- this pointer points to some indeterminate location. You could be just lucky in that writing values to this "random" location works and won't result in a crash. I'd personally call it unlucky because this hides a bug in your program. undefined behavior just means anything can happen, including a "correct" program execution.
malloc() will dynamically request some usable memory and return a pointer to that memory, so after the malloc(), you know i points to 4 bytes of memory you can use. If malloc() fails for some reason (no more memory available), it returns NULL -- your program should test for it before writing to *i.
All that said, of course the program CAN work without malloc(). You could just write
char i[4];
and i would be a local variable with room for 4 characters.
Final side note: sizeof(char) is defined to be 1, so you can just write i = malloc(4);.
Unfortunately, "runs fine" criterion proves nothing about a C program. Great deal of C programs that run to completion have undefined behavior, which does not happen to manifest itself on your particular platform.
You need special tools to see this error. For example, you can run your code through valgrind, and see it access uninitialized pointer.
As for the malloc, you do not have to use dynamic buffer in your code. It would be perfectly fine to allocate the buffer in automatic memory, like this:
char buf[4], *i = buf;
You have to allocate space for memory. In the example below, I did not allocate for memory for i, which resulted in a segmentation fault (you are trying to access memory that you don't have access to)
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
char *i;
strcpy(i, "hello");
printf("%s\n", i);
return (0);
}
Output: Segmentation fault (core dumped)
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
char *i;
/*Allocated 6 spots with sizeof char +1 for \0 character*/
i = malloc(sizeof(char) * 6);
strcpy(i, "hello");
printf("%s\n", i);
return (0);
}
Result: hello
Malloc allows you to create space, so you can write to a spot in memory. In the first example, "It won't work without malloc" because i is pointing to a spot in memory that doesn't have space allocated yet.
Good evening everybody, I am learning C++ on Dev C++ 5.9.2, I am really novice at it. I intentionnally make my programs crash to get a better understanding of bugs. I've just learned that we can pass a char string to a function by initializing a pointer with the address of the array and that was the only way to do it. Therefore we should always pass to the function the size of that string to handle it properly. It also means that any procedure can run with a wrong size passed in the argument line hence I supposed we could read farther than the allocated memory assigned to the string.
But how far can we do it? I've tested several integers and apparently it works fine below 300 bytes but it doesn't for above 1000 (the program displays characters but end up to crash). So my questions are :
How far can we read or write on the string out of its memory range?
Is it called an overflow?
How does the program detect that the procedure is doing something unlegit?
Is it, the console or the code behind 'cout', that conditions the shutting down of the program?
What is the condition for the program to stop?
Does the limit depend on the console or the OS?
I hope my questions don't sound too trivial. Thank you for any answer. Good day.
#include <iostream>
using namespace std;
void change(char str[])
{
str[0] = 'C';
}
void display(char str[], int lim)
{
for(int i = 0; i < lim; i++) cout << str[i];
}
int main ()
{
char mystr[] = "Hello.";
change(mystr);
display(mystr, 300);
system("PAUSE");
return 0;
}
The behavior when you read past the end of an array is undefined. The compiler is free to implement the read operation in whatever way works correctly when you don't read beyond the end of the buffer, and then if you do read too far - well, whatever happens is what happens.
So there are no definite answers to most of your questions. The program could crash as soon as you read 1 byte too far, or you could get random data as you read several megabytes and never crash. If you crash, it could be for any number of reasons - though it likely will have something to do with how memory is managed by the OS.
As an aside, the normal way to let a function know where a string ends is to end it with a null character rather than passing a separate length value.
program in c language
void main()
{
char *a,*b;
a[0]='s';
a[1]='a';
a[2]='n';
a[3]='j';
a[4]='i';
a[5]='t';
printf("length of a %d/n", strlen(a));
b[0]='s';
b[1]='a';
b[2]='n';
b[3]='j';
b[4]='i';
b[5]='t';
b[6]='g';
printf("length of b %d\n", strlen(b));
}
here the output is :
length of a 6
length of b 12
Why and please explain it.
thanks in advance.
You are assigning to pointer (which contains garbage) without allocating memory. What you are noticing is Undefined Behavior. Also main should return an int. Also it does not make sense to try and find the length of an array of chars which are not nul terminated.
This is how you can go about:
Sample code
When you declare any variable it comes with whatever it had in memory previously where your application is running, and since pointers are essentially numbers, whatever number it had referenced to some random memory address.
Then, when setting a[i], the compiler interprets that as you want to step sizeof(a) bytes forward, thus, a[i] is equal to the address (a + i*1) (1 because chars use one byte).
Finally, C-strings need to be NULL terminated (\0, also known as sentinel), and methods like strlen go over the length of the string until hitting the sentinel, most likely, your memory had a stray 0 somewhere that caused strlen to stop.
Allocate some memory and terminate the strings then it will work better
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
void main(){
char *a=malloc(10);
char *b=malloc(10);
if(a){
a[0]='s';
a[1]='a';
a[2]='n';
a[3]='j';
a[4]='i';
a[5]='t';
a[6]=(char)0;
printf("length of a %d\n", (int)strlen(a));
}else{
printf("Failed to allocate 10 bytes\n" );
}
if(b){
b[0]='s';
b[1]='a';
b[2]='n';
b[3]='j';
b[4]='i';
b[5]='t';
b[6]='g';
b[7]=(char)0;
printf("length of b %d\n", (int)strlen(b));
}else{
printf("Failed to allocate 10 bytes\n" );
}
free(a);
free(b);
}
Undefined behavior. That's all.
You're using an uninitialized pointer. After that, all bets are off as to what will happen.
Of course, we can attempt to explain why your particular implementation acts in a certain way but it'd be quite pointless outside of novelty.
The indexing operator is de-referencing the pointers a and b, but you never initialized those pointers to point at valid memory. Writing to un-initialized memory triggers undefined behavior.
You are simply "lucky" (or unlucky, it depends on your viewpoint) that the program doesn't crash, that the pointer values are such that you succeed in writing at those locations.
Note that you never write the termination character ('\0') to either string, but still get the "right" value from strlen(); this implies that a and b both point at memory that happens to be full of zeros. More luck.
This is a very broken program; that it manages to run "successfully" is because it's behavior is undefined, and undefined clearly includes "working as the programmer intended".
a and b are both char pointers. First of all, you didn't initialise them and secondly didn't terminate them with NULL.
I am learning how to use pointers, so i wrote the below program to assign integer values in the interval [1,100] to some random locations in the memory.
When i read those memory locations, printf displays all the values and then gives me a segmentation fault. This seems an odd behavior, because i was hoping to see either all the values OR a seg fault, but not both at the same time.
Can someone please explain why i got to see both?
Thanks. Here is the code
#include <stdio.h>
#include <stdlib.h>
int main()
{
char first = 'f';
char *ptr_first = &first;
int i=1;
for(i=1;i<101;i++)
*(ptr_first+i) = i;
for(i=1;i<101;i++)
printf("%d\n", *(ptr_first+i));
return EXIT_SUCCESS;
}
Not odd at all. You are using your variable first, which is on the stack. What you essentially do is happily overwriting the stack (otherwise known from buffer overflows on the stack) and thus probably destroying any return address and so on.
Since main is called by the libc, the return to libc would cause the crash.
You're accessing memory past beyond that assigned to first. It is just one character, and, through the ptr_first pointer, you're accessing 100 positions past this character to unreserved memory. This may lead to segfaults.
You have to ensure the original variable has enough memory reserved for the pointer accesses. For example:
char first[100];
This will convert first in an array of 100 chars (basically a memory space of 100 bytes that you can access via pointer).
Note also that you're inserting int into the char pointer. This will work, but the value of the int will be truncated. You should be using char as the type of i.
since ptr_first pointer is pointing to a char variable first. Now when you are incrementing ptr_first, so incremented memory address location can be out of process memory address space, thats why kernel is sending segmentation fault to this process.