Assign part of a string to a char array

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Edit: After asking this question I was informed about the question How to initialize a char array without the null terminator? which is almost identical. (I'm building a network packet as well.)

I'll keep this question open anyway, because I'm just assigning and not initializing.

I have a fixed size array that I would like to assign a fixed text to:

char text[16]; text = "0123456789abcdef";

Of course this doesn't work because the right hand side contains the null terminator.

error: incompatible types in assignment of 'const char [17]' to 'char [16]'

The text is human-readable, so I would prefer to keep it in one piece, i.e. not write {'0', '1', ...}.

Can I make the assignment work somehow?

By the way, I only have a few hundred bytes of RAM, so preferably (but second to the human-readability requirement) the solution shouldn't use twice the RAM for a temporary copy or something like that.

In C, strcpy(text, "0123456789abcde"); /* text has space for 15 "real" characters and the zero terminator */ Note that strcpy(text, "0123456789abcdef"); is an error! — pmg
– pmg, Commented Jul 14, 2020 at 11:36
So the question is about how to store 17 bytes in 16 byte array? — Captain Trojan
– Captain Trojan, Commented Jul 14, 2020 at 11:41
@AndreKR Then see How to initialize a char array without the null terminator? TLDR: char text[16] = "0123456789abcdef"; (note that won't work in C++) — Andrew Henle
– Andrew Henle, Commented Jul 14, 2020 at 12:17

fsquirrel · Accepted Answer · 2020-07-14 11:50:30Z

If you write in C:

char text[16]; strncpy(text, "0123456789abcdef", sizeof(text));

Note that text will not have the null-terminator and won't be compatible with the standard C functions like strlen. If the text if human-readable, I recommend to include the terminator. It is only one character but it will make your life much easier.

Example:

char text[17]; strncpy(text, "0123456789abcdef", sizeof(text));

aafulei · Accepted Answer · 2020-07-14 12:32:48Z

First of all, in case you have any misunderstanding -- "0123456789abcdef" is a string literal. Its type is const char [17], not const char [16], because this string literal has a null terminator \0 in the end.

Then I assume you do mean assignment not initialization. They are different.

I can think of multiple ways to do the assignment. You may choose based on your needs.

#include <cstddef> #include <cstring> #include <iostream> #include <string> #include <string_view> using std::size_t; template<size_t N> void print(const char (&a)[N]) { for (size_t i = 0; i != N; ++i) putchar(a[i]); putchar('\n'); } void foo1() { char text[16]; std::memcpy(text, "0123456789abcdef", sizeof text); print(text); } void foo2() { char text[16]; std::string("0123456789abcdef").copy(text, sizeof text); print(text); } void foo3() { char text[16]; std::string_view("0123456789abcdef").copy(text, sizeof text); print(text); } // programmer needs to make sure access with src is valid template<size_t N> void assign(char (& dest)[N], const char * src) { for (size_t i = 0; i != N; ++i) dest[i] = *src++; } void foo4() { char text[16]; assign(text, "0123456789abcdef"); print(text); } int main() { foo1(); // 0123456789abcdef foo2(); // 0123456789abcdef foo3(); // 0123456789abcdef foo4(); // 0123456789abcdef return 0; }

Some remarks:

std::memcpy is the traditional memory copy way from C
std::string_view is a light-weight view on a string, introduced in C++17
Or you may write your own function like assign() -- with a template you can tell the size of an array at compile time. Depending on your needs, you may decide whether or not to implement bounds checking.

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