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Segmentation Fault doesn't come up immediately after accessing out-of-bound memory
(5 answers)
Closed 7 years ago.
What's stored after the end of character array? I was assuming there would be some random garbage but it did not print anything after the end while looping 10 times.
char a[] = "Pencil";
int i;
for (i = 0; i < 10; i++)
{
printf("%c", a[i]);
}
so the character array a has the size of 7. And for loop looped until 10th position which are 3 more values looped through. But it did not print anything or Error. What's going on here?
Accessing beyond the end of an array in C is undefined behavior. Your program could continue running unchanged or it could crash horrifically depending on what is stored past the end of the array. The compiler makes no guarantees about what is stored there - it could be useless memory or it could be critical to your program.
Accessing an array element beyond the array-length invokes undefined behavior which means anything could happen.
Related
This question already has answers here:
How dangerous is it to access an array out of bounds?
(12 answers)
Why doesn't my program crash when I write past the end of an array?
(9 answers)
Array index out of bound behavior
(10 answers)
No out of bounds error
(7 answers)
Closed 3 years ago.
I was making practises on the logic of arrays in c and my thought on the array length declaration was unformattable if you declare an array length to be 10 integers, that array could not keep 20 integers in memory but when I tested it I saw that I was completely wrong
int main(){
int i;
int arr[10];
for (i = 0;i<20;i++){
arr[i] = i;
}
for (i = 0;i<20;i++){
printf("%d \n",arr[i]);
}
}
I was expecting to see 10 printed numbers but it prints out 20 could someone explain how is it possible?
C and C++ don't have explicit bounds checking on array sizes. When you read/write past the end of an array, you invoke undefined behavior.
With undefined behavior, your program may crash, it may output strange results, or (as in your case) it could appear to work properly. Also, making a seemingly unrelated change such as adding an unused local variable or adding a printf for debugging can change how UB manifests itself.
Just because a program may crash doesn't mean it will.
This question already has answers here:
Why doesn't my program crash when I write past the end of an array?
(9 answers)
Closed 7 years ago.
I am getting confused with an array code.
According to me the program should raise an error but it's working fine. The code :
#include<stdio.h>
#include<conio.h>
void main()
{
int a[1],n,i;
clrscr();
printf("Enter the length");
scanf("%d",&n);
for( i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
for(i=0;i<n;i++)
{
printf("%d ",a[i]);
}
getch();
}
Here the array size is 1 but when I enter the length 5 then it works fine : the program show all 5 elements that i have entered.
This is the output screen.
Accessing array out of bounds causes undefined behavior. Anything can happen including the outcome you are observing. In this case you are overwriting some objects stored after the array. They are just not used in this particular case and your program doesn't crash.
Such bugs are really hard to debug. It works fine now, but might start to fail, for example, when different compiler is used. Memory analyzer can help detecting such bugs. It will detect some invalid memory accesses, even if they do not cause a crash.
When you have defined int a[1], only space for one int is allocated on the stack. Any access beyond array bound causes undefined behavior. Therefore, the code is wrong according to the C standard.
In your case, the program is accessing some space beyond the array and by the chance of luck you didn't end up yourself with a segmentation fault.
This question already has answers here:
Array index out of bound behavior
(10 answers)
Closed 8 years ago.
I was trying to test what output I would get when I tried to print the array that the index goes out of bound.
The code:
#include <stdio.h>
void main()
{
int arr[] = { 3, 4, 5 };
for (int i = 0; i < 5; i++)
{
if (arr[i] == 0)
{
printf("Breaking out of the loop.");
break;
}
else
printf("%i\n", arr[i]);
}
getchar();
}
When I run this code, the output is:
3
4
5
-858993460
3997200
I expected it to print out "Breaking out of the loop" and break out of the loop and terminate.
I truly have no idea how it even printed those numbers.
Any idea what those numbers mean?
P.S. I am sorry if this is a stupid question, I am quite new to C.
Memory out of bounds of an array, or dynamically allocated memory, doesn't belong to you, and its content is indeterminate. Accessing arrays or memory out of bounds leads to undefined behavior. Just don't do it.
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Closed 9 years ago.
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I read this paragraph in c the complete reference book
C has no bounds checking on arrays. You could overwrite either end of
an array and write into some other variable's data or even into the
program's code. As the programmer, it is your job to provide bounds
checking where needed. For example, this code will compile without
error, but it is incorrect because the for loop will cause the array
count to be overrun.
#include <stdio.h>
int main(){
int count[10], i;
/* this causes count to be overrun */
for(i=0; i<15; i++) count[i] = i;
for(i=0; i<15; i++) printf("%d ",count[i]);
return 0;
}
and when I try the code it gives me
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
And then a run time error was displayed on the screen. The error was
array.exe has stopped working
My question is : What is the type of that error? And does it mean that my IDE checks the bounds of the array?
C doesn't check array idexing, it does only simple pointer-arithmetic. That is, the standard defines the operation array[i] as array + i.
Depending on the resulting memory address of that computation, many things can occur:
The address points to the location of a local variable of the function: You modify the local variable. For example:
int main()
{
int array[10];
int a = 0;
array[11] = 22;
printf(a%i,a); /* Prints 22 */
}
Remember that the point here is that buffer overrun has undefined behaviour. My example could not work, because the compiler is free to reorder the variables layout during compilation, for memory-aligment and optimising purposes.
The address points to the location of a global variable (Thats a very big indexing/jump, but can occur): The effect is the same as in the local variable case.
The address points to the location where the subroutine stores the return address: Many architectures stores the return address of a function in a register, but if the architecture stores it in the stackframe of the function... WOW. TRY TO DEBUG THAT!!! :)
The address goes out of the memory space of the process: Modern operative systems always check memory accesses to prevent this, so when this happens the OS kicks you in the ass and throws an exception.
Finally note that things are only applicable in release compilation. In debug mode the compiler adds a lot of code to check this kind of things to throw (Provide) a comprehensible and easy-debuggeable exception.
For example: When allocating dynamic arrays, windows debug-heap first fills the memory space that will be used with a flag which says that memory space is ready to use, but it doesn't contains any data: Thats the hexspeak 0xBADF00D. After this, malloc retrieves the memory location, adding hexspeaks around the array to provide bounds checking.
See this article for a complete explanation.
The moment the first loop reaches i=10, you enter the realm of undefined behaviour (since you're writing past the end of count). From that point on, anything can happen.
Your IDE had nothing to do with it. The OS killed your program because it tried to access memory that didn't belong to it.
This type of error is usually known as an access violation.
That's fine - and the point of the article you have read...
What you are supposed to do is limit the for loop so that it only runs from 0 to 9.
Or, if you really need 15 items, expand the array to take them...
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Array index out of bound in C
Can a local variable's memory be accessed outside its scope?
C no out of bounds error
I am trying out this code snippet,
#include <stdio.h>
main(){
int a[2],i;
a[5] = 12;
for(i=0;i<10;i++){
printf("%d\n", a[i]);
}
return 0;
}
It gives me output :
1053988144
32767
0
3
0
12
-1267323827
32716
0
0
Why a[5] is accessible ? Shouldn't it through RunTime Error?
C doesn't have bounds-checking on array access, so no it shouldn't. There is nothing in the language stopping you from attempting to read every (virtual) address you can imagine, although often the operating system and/or computer itself will protest, which might generate some kind of exception. Note that just because you "can" (it compiles and run, seemingly without any ill effects), that doesn't mean the program is valid or that it "works". The results of invalid memory accesses are undefined behavior.
int a[2]; means "allocate memory that is 2 * sizeof(int)"
a[5] is syntactic sugar for *(a + 5), which point to an area of memory a + (5 * sizeof(int)). So 3 * sizeof(int) past the end of your array. Where is that? Who knows?
Some languages do bound checking, and I have heard of some C compilers that can do it as well, but most do not. Why not do bound checking? Performance. And performance is a major reason for using C in the first place. But that's OK, because C programmers are good programmers and never go beyond the bounds of an array. (hopefully)
No control is performed on index bounds. This is simply the way C works.