Conversion from string literal to char *

Question

I am trying to make a c function work in some publicly available linear algebra code.

the publicly available prototype is…

int ilaenv_(int *, char *, char *, int *, int *,int *, int *);

The publicly available code has the function call…

nb = ilaenv_(&c__1, "DGEQRF", " ", m, n, &c_n1, &c_n1);

where m, n, c_1, and c_n1 are integers,

The error message is.

C++ 11 does not allow conversation from string literal to char *.

I did not create the code, but downloaded it from the LAPACK site. I hesitate to make too many changes to publicly available code that supposedly works, for fear of introducing errors. However, this error is showing up on a number of functions in the program that I am working on.

How can I resolve this?

You can just cast constness but I don't know if that can be good. — Iharob Al Asimi, Mar 04 '15 at 17:56
Does it really say `conversation`? Please use the actual error strings. And fix the title. — Eugene Sh., Mar 04 '15 at 17:57
Not a strict duplicate, but this should have some helpful information: http://stackoverflow.com/questions/1704407/what-is-the-difference-between-char-s-and-char-s-in-c/1704433#1704433 — Adam, Mar 04 '15 at 17:57
Yes please post the exact error message, copy and paste, it's not possible that it says `conversation`. — Iharob Al Asimi, Mar 04 '15 at 17:58
I guess you can't talk about string literals and char* in c++11. Freaken passive-aggressive language... — thang, Mar 04 '15 at 18:00
Oops. one of those "spell checkers" got me again. "conversation" should be conversion. So the exact message is .... "C++ 11 does not allow conversion from string literal to char * — K17, Mar 05 '15 at 02:31

score 8 · Answer 1 · answered Mar 05 '15 at 10:16

8

Your function takes "char *", not "const char *". String literals can be assigned only to "const char *".

answered Mar 05 '15 at 10:16

senfen

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21

score 6 · Answer 2 · answered Mar 04 '15 at 18:01

6

To be safe, you could create char arrays initialised to the same values, for example:

char dgeqrf[] = "DGEQRF";
char space[] = " ";

Or you could check the source code of the function; if it doesn't actually modify the contents of those arrays you could change the arguments to const char *.

answered Mar 04 '15 at 18:01

Arkku

41,011
10
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84

Thank you. This looks like the direction to go. – K17 Mar 05 '15 at 02:33
Creating the character string successfully removed the error. – K17 Mar 05 '15 at 02:59

score 1 · Accepted Answer · answered Nov 08 '17 at 16:10

1

Sorry for being so late, I just bumped into this problem. You can try this,

string dgeqrf = "DGEQRF";

One problem with using "const char *" is code like this,

string formatString;
switch (format)
{
    case FORMAT_RGB:  formatString = "RGB"; break;
    case FORMAT_RGBA: formatString = "RGBA"; break;
    case FORMAT_TP10: formatString = "TP10"; break;
    default:          formatString = "Unsupported"; break;
}

If you declare formatString as a "const char *", this won't work.

answered Nov 08 '17 at 16:10

Albertus

56
4

re. "if you declare `formatString` as a `const char *` this won't work" – the code shown here will work just fine with `const char *`, but the potential problem is that the function takes `char *` not `const char *` so we don't know if it is going to modify the contents of the string, and modifying the string literals is not permitted. Hence the need to have a modifiable array, either by conversion from `string` or as a `char` array instead of pointer (as per my answer). _Unless_ you need to assign the contents dynamically (like in this answer), the array is simpler. – Arkku Nov 15 '17 at 09:14

Conversion from string literal to char *

3 Answers3