I want to assign one char array to another, but the following does not work.
unsigned char myval1[6] = {0x01, 0x01, 0x01, 0x01, 0x01, 0x01};
unsigned char myval2[6];
myval2 = &myval1[0];
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What you are trying to do gives compiler error because, an array variable here myval1,myval2 are always constant type, so you cannot modify base address of an array which you are actually doing in your solution. You can follow any of the answers given below for copying your arrays.prince– prince2014-04-07 09:47:41 +00:00Commented Apr 7, 2014 at 9:47
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7 Answers
C-Array are not copy-assignable.
You have to do the copy yourself (using std::copy (C++) or memcpy (C)).
In C++11, you may use std::array instead.
Comments
This is one thing you cannot do with an assignment. One possibility is
memcpy( myval2, myval1, sizeof(myval2) );
Comments
In C you an not assign the array. You should use memcpy() function to copy array.
memcpy(myval2, myval1, sizeof(myval2));
Comments
You might try something like this..
unsigned char myval1[6] = {0x01, 0x01, 0x01, 0x01, 0x01, 0x01};
unsigned char *myval2;
myval2 = &myval1[0];
1 Comment
laune
Probably not what OP had in mind. It creates an alias for myval1 and does not copy any values.
No, you cannot assign arrays in C.
You must manually copy each array element. The standard library function to do this is memcpy():
memcpy(myval2, myval1, sizeof myval2);
The final argument is the number of bytes to copy, and using sizeof is very much recommended. Note that if myval1 is smaller than myval2, the above will be undefined behavior since it will read beyond the bounds of myval1. You must be careful.
Comments
A good idiom for this is:
memcpy(&myval2, &myval1, sizeof myval2);
By pairing &x and sizeof x you ensure that you always copy the right number of bytes, even if you make a mistake elsewhere with variable names.
The version suggested in the other posts works anyway because myval2 is an array, so (void *)myval2 == (void *)&myval2.
Comments
try this
unsigned int myval1[6] = {0x01, 0x01, 0x01, 0x01, 0x01, 0x01};
unsigned int myval2[6];
unsigned int* add = &myval1[0];
cout << *add << endl;
2 Comments
Moha the almighty camel
and how does this copy the values of
myval1 to maval2 ?
Sks Sapare
there is no myval2 here, you need to use variable add, add++ to access next values. Yes variable add address will be pointing to the starting address of myval1 so *add will print 1 ; *add++ = 1 [myval1[1]] ; *add++ = 1 [myval1[2]]