This question already has answers here:
With arrays, why is it the case that a[5] == 5[a]?
(20 answers)
Closed 9 years ago.
I am using Ubuntu 12.04lts with the GCC compiler. This program gives the result 10. Could you anybody please describe why this program gives the result like this?
#include <stdio.h>
void main(void)
{
int arr[1] = {10};
printf("\n%d\n\n", 0[arr]);
}
arr[0] gets internally expanded to *(arr+0). Similarly 0[arr] gets expanded to *(0+arr) which points to the same thing. Hence you see 10.
In general for an array or a pointer a, a[b] always means *(a+b) where a is the starting address of the array or pointer and b is the offset. Thus, a[b] and b[a] are equivalent.
below line means arr is int type array and it has size 1 and it is initialized with 10 ie index 0 has 10
int arr[1] = {10};
then next line printf statement printing the value of arr at index 0.
printf("\n%d\n\n",0[arr]);
Related
This question already has answers here:
With arrays, why is it the case that a[5] == 5[a]?
(20 answers)
Closed 1 year ago.
I saw such a piece of C code:
int main()
{
static int a[] = {7,8,9};
printf("%d", 2[a] + a[2]);;
return 0;
}
What does 2[a] mean here?
a[b] and b[a] are 100% equivalent in C. What you have there is a very unidiomatic way of writing a[2].
By way of more complete explanation, array subscript notation a[b] is also 100% equivalent to *(a + b), which may make the reason it works both ways clearer.
This question already has answers here:
What does 1[d=b] mean?
(3 answers)
Closed 5 years ago.
I came across a code as in below
#include <stdlib.h>
int main(){
char a[]="0123456789";
printf("%s\n",a+6[a]-2[a]);
return 0;
}
Output
456789
How does the calculation of a+6[a]-2[a] happens in printf?
Why giving just 6[a] in printf doesn't work?
printf("%s\n",6[a]);
Well, a statement like
a+6[a]-2[a]
can be re-written as
&(a[ a[6] - a[2] ])
which is simply,
use the value of a[6] (type, int) as the index in the first case
use the value of a[2] (type int) as the RHS.
The result, is a pointer, is passed to printf() as an argument to %s conversion specification.
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:
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:
With arrays, why is it the case that a[5] == 5[a]?
(20 answers)
Closed 8 years ago.
#include <stdio.h>
int main()
{
int a=3, b = 6;
printf(&a["Hi!Hello! %s\n"], &b["Mnnit/Softathalon"]);
printf(&a["WHAT%c%c%c %c%c %c !\n"], 1["this"],
2["beauty"],0["tool"],0["is"],3["sensitive"],4["CCCCCC"]);
return 0;
}
output:
Hello! Softathalon
That is C !
Why is this the output? Can anyone explain different format specifier in it?
For any array T arr[N], the expression arr[i] is equivalent to *(arr + i).
Because the addition is commutative in the latter expression, you can also write this as *(i + arr), and hence as i[arr].
In particular, arr[3] and 3[arr] denote the same thing.
It's one of those "curiously funny things you can do in C", but it should go without saying that serious code should never actually use such a construction.