C++ char array: error while copying data

Question

I have an array:

CHAR m_manuf[256];

I am trying to copy a value to this array as:

m_manuf = "abacus";  //This shows error

I also tried this variation:

char abc[256] ="abacus";
m_manuf = abc; //Shows error as left value must be l-value

You can't assign to an array, but you can copy to it. But if you're programming in C++, why don't you use std::string? — Some programmer dude
– Some programmer dude, Commented Jan 25, 2016 at 7:20
You can also do something like char * m_manuf = "abacus"; probably. — CollinD
– CollinD, Commented Jan 25, 2016 at 7:21
@CollinD It's generally a bad idea to have a non-const pointer to a string literal. — Some programmer dude
– Some programmer dude, Commented Jan 25, 2016 at 7:28

Akhil V Suku · Accepted Answer · 2016-01-25 07:25:41Z

2

You cant' copy an array like that, instead of you can do,

CHAR    m_manuf[256];
strcpy(m_manuf,"abacus" );

Or

char * m_manuf = "abacus";

Or

char abc[256] ="abacus";
strcpy(m_manuf,abc );

Note : The better way to handle char arrays are using std::string,

answered Jan 25, 2016 at 7:25

Akhil V Suku

9122 gold badges13 silver badges36 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Roger · Accepted Answer · 2016-01-25 07:30:40Z

0

A constant char array is good enough for you so you go with,

string tmp = "abacus";
char *new = tmp.c_str();

Or you need to modify new char array and constant is not ok, then just go with this

char *new = &tmp[0];

answered Jan 25, 2016 at 7:30

Roger

1,2358 silver badges16 bronze badges