can you help me with this code, i am able to copy arr into arr2(from index 1 and further, i cannot figure index 0 till now), but some others can't. Can you tell if anything's wrong here?
#include <stdio.h>
int main(){
char arr[] = "char_arr_one";
char arr2[] = {arr};
printf("%s\n%s\n", arr, arr2);
int i;
for(i=0;i<13;i++){
printf("%c %c\n", arr[i], arr2[i+1]);
}
}
this is working on my system!
The C standard does not permit to assign an array to another array:
char arr[] = "char_arr_one";
char arr2[] = {arr}; // This does not work.
Why it works on a specific implementation as you´d stated, I don´t know, but generally it should not be possible to compile that code without at least one warning.
So maybe you disabled or even ignored warnings?
One link to that context, here: Why should I always enable compiler warnings?
Nonetheless, this doesn´t bring you the desired result of copying the string "char_arr_one" into arr2.
If you want to store the string, stored inside of arr, in arr2 you can use strcpy in the header string.h:
char arr[] = "char_arr_one";
char arr2[13]; // You need to provide the amount of elements, at least
// as much as are required to store the string inside arr
// the null character.
strcpy(arr2,arr); // Copies the string in arr into arr2.
Note, that you need to specify the elements of arr2, which need to be at least as much as are required to store the string of "char_arr_one" plus the terminating null character \0, when defining arr2. In this case, arr2 needs to have at least 13 char objects.
You could also "automatically" detect the size of arr by using the sizeof operator:
char arr[] = "char_arr_one";
char arr2[sizeof(arr)]; // Automatically detects the size of `arr` and provides
// it for specify the required amount of elements for storing
// the string in arr + the null character.
strcpy(arr2,arr); // Copies the string in arr into arr2.
Beside that, the third argument of arr2[i+1] inside the printf call will get you into Undefined Behavior. At the last iteration, it would print something what lies beyond the array of arr2. So change that to arr2[i].
The corrected code shall be:
#include <stdio.h>
#include <string.h>
int main(){
char arr[] = "char_arr_one";
char arr2[sizeof(arr)];
strcpy(arr2,arr);
printf("%s\n%s\n",arr,arr2);
for(int i = 0; i < 13; i++){
printf("%c %c\n", arr[i], arr2[i]);
}
return 0;
}
Output:
char_arr_one
char_arr_one
c c
h h
a a
r r
_ _
a a
r r
r r
_ _
o o
n n
e e
in C you can't assign array to array.
You need to copy it yourself
char arr[] = "char_arr_one";
char arr2[sizeof(arr)];
memcpy(arr2, arr, sizeof(arr);
or
for(size_t index = 0; index < sizeof(arr); index++)
{
arr2[index] = arr[index];
}
If I make your example a bit less trivial (by adding another variable) the result is obvious. It does not work.
#include <stdio.h>
int main(){
char arr[] = "char_arr_one";
char another[] = "it does not work for sure";
char arr2[] = {arr};
printf("%p %p\n", (void *)arr, (void *)arr2);
printf("%s\n%s\n", arr, arr2);
int i;
for(i=0;i<13;i++){
printf("0x%02hhx - '%c' 0x%02hhx, '%c'\n", arr[i], arr[i],
arr2[i], (arr2[i] >= 32 && arr2[i] <= 127) ? arr2[i] : ' ' );
}
}
And the result is:
0x7fff4a68b84f 0x7fff4a68b82f
char_arr_one
Oit does not work for sure
0x63 - 'c' 0x4f, 'O'
0x68 - 'h' 0x69, 'i'
0x61 - 'a' 0x74, 't'
0x72 - 'r' 0x20, ' '
0x5f - '_' 0x64, 'd'
0x61 - 'a' 0x6f, 'o'
0x72 - 'r' 0x65, 'e'
0x72 - 'r' 0x73, 's'
0x5f - '_' 0x20, ' '
0x6f - 'o' 0x6e, 'n'
0x6e - 'n' 0x6f, 'o'
0x65 - 'e' 0x74, 't'
0x20 - '' 0x20, ' '
arr accessing its elements by reading out of bounds arr2 which is 1 byte long and it is before the arr as they are stored on the stack and newer elements have lower addresses.i am able to copy arr into arr2(from index 1 and further, i cannot figure index 0 till now), but some others can't. .....
On what basis you are sure about this? I do not see any code which is copying arr to arr2 from index 1 and further.
Look at the gcc compiler warning on the statement which is, as per you, copying arr to arr2 from index 1:
<source>:5:24: warning: initialization of 'char' from 'char *' makes integer from pointer without a cast [-Wint-conversion]
5 | char arr2[] = {arr};
Meaning of this warning -
The array name arr decays to pointer and the pointer is converted to integer and assigned to the first element of arr2 (element at 0th index of array arr2).
Note that if the char is signed then the result of this assignment will be implementation defined1).
The way the arr and arr2 placed in stack on your machine, you may be getting expected output but on a different platform/architecture you may get different result. Note that C language standards do not specify how a function local variables should be placed in stack. The standards even do not have a single mention of memory segments (stack, heap etc.) in it. These things are completely dependent on the underlying platform/architecture. Standards only talk about the scope and life of variables.
Can you tell if anything's wrong here?
Since you have omit the dimension of arr2, the compiler computes it based on the size of the initializer. The initializer of arr2 is having only 1 element. So, the size of arr2 will be 1 i.e. arr2 is a char array with just one element. Here you accessing the arr2 beyond its size:
for(i=0;i<13;i++){
printf("%c %c\n", arr[i], arr2[i+1]);
^^^^^^^^^
Accessing an array beyond its size is undefined behavior. An undefined behavior includes it may execute incorrectly (either crashing or silently generating incorrect results), or it may fortuitously do exactly what the programmer intended.
1) From C Standards#6.2.5p15
15 The three types char, signed char, and unsigned char are collectively called the character types. The implementation shall define char to have the same range, representation, and behavior as either signed char or unsigned char.45)
warning: initialization of ‘char’ from ‘char *’ makes integer from pointer without a cast [-Wint-conversion](even with no warning options explicitly enabled). You should heed compiler warnings — the compiler knows more about C than you do.