convert char* to const char*

Hi,
how do i convert char* to const char* in the following example?

Code:

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        UCHAR ucar[3];
        ucar[0]='a';
        ucar[1]='b';
        ucar[2]='c';
       
            //convert char* to  const char*
            //const char* car;

        printf(car);

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-------------------------Method 1------------------------
Declare your array as CHAR (recommended)

-------------------------Method 2-------------------------
Cast to (char*) ---> printf((char*)ucar);

You don't need to cast something from type to const type because if you pass type to a function expecting a const type, it will automatically be upgraded.

If you have a const type and need to pass it to a function that isn't const-correct, you can cast it like Elysia suggested.

Unfortunately, this is unsigned char* and printf expects char*, thus causing a compile error in C++ and a warning in C.

Quote:

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Originally Posted by Elysia

Unfortunately, this is unsigned char* and printf expects char*, thus causing a compile error in C++ and a warning in C.

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Shouldn't matter since OP isn't dealing with them arithmetically and neither does printf() with the format string. I am not sure what OP is trying to accomplish other than buffer overrun though.

Shouldn't matter, but does matter since it invokes a warning. Thus, a cast or change of types would be the best solution - compilers output warnings for a reason and it's best to satisfy the compiler by solving them.

The lack of a \0 in the string is another problem to be addressed.

I just saw that now. That would be why we usually do

Code:

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UCHAR ucar[] = "abc";

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Or

Code:

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UCHAR uchar[4];
strcpy((char*)uchar, "abc");

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Though I'm not suer if "abc" will be considered just const char* and require a cast to unsigned const char*.
Usually, unsigned char is typically used for buffers; when you use strings, normal char is what you use.