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.
Related
This question already has answers here:
Why doesn't my program crash when I write past the end of an array?
(9 answers)
How dangerous is it to access an array out of bounds?
(12 answers)
Closed 5 years ago.
Here is my code:
#include<stdio.h>
int main()
{
int i, list[1];
list[0]=1;
list[1]=2;
list[2]=7;
list[55]=70;
i=sizeof list;
printf("%d %d %d %d %d Size of array is %d",list[0],list[1],list[2],list[3],list[55],i);
return(0);
}
It returns "1 2 7 4 70 Size of array is 4". Why can i assign, say 55 to list[55]. list[55] should not exist as I only gave the array list enough memory for 1 integer, right? In addition shouldn't this give me an error as list[3] doesn't exist? and if for some reason i am changing the size of the array why isn't the size 56? It comes out as 4.
So what is happening to give me the output i got?<--{main question}
[As i don't want to create a separate thread for a related question, why when i code int list[0]; the program crashes, if i am somehow changing the size from 1 to 4 shouldn't I be able to change the size from 0 to 4?]
Thanks for your help, I know this probably a stupid or obvious question.
This question already has answers here:
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.
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:
With arrays, why is it the case that a[5] == 5[a]?
(20 answers)
Closed 8 years ago.
Can anyone help me out, I am not getting how gcc is compiling the below statement and printing its output.:-
printf("%d",7["sunderban"]);
C allows to access array's elements in two ways (see Accessing arrays by index[array] in C and C++ and the answers) :
int v[5];
// 1)
v[2] = 33;
// 2)
2[v] = 44;
So, what happens in your case is that you access the 8th element of the string, and it is interpret as int by the printf.
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.