C Programming Language :: assigning const char* to char*


	assigning const char* to char* Hi, This is a very simple question but I couldn't find it in your FAQ. I'm using VC++ and compiling a C program, using the /TC flag. I've got a function for comparing two strings int strspcmp(const char * s1, const char * s2) { char* pS1 = s1; char* pS2 = s2; ... ... } As you can see, I assign the two const ptrs to some non-const ptrs to do some ptr arithmetic, like incrementing the ptrs to compare the strings. But I'm getting warning C4090 complaining about the assignment. I thought I was doing the standard thing for ptr arithmetic. What am I doing wrong? How do I get rid of the warning? Peithon said: > Hi, > This is a very simple question but I couldn't find it in your FAQ. > I'm using VC++ and compiling a C program, using the /TC flag. > I've got a function for comparing two strings > int strspcmp(const char * s1, const char * s2) Although it has nothing to do with your question, I should perhaps point out that the name strspcmp is reserved. str_spcmp is not, however. > { > char* pS1 = s1; > char* pS2 = s2; > ... > ... > } > As you can see, I assign the two const ptrs to some non-const ptrs to > do some ptr arithmetic, > like incrementing the ptrs to compare the strings. You don't need to do that, and in fact it's a bad idea. The const char * syntax doesn't mean "this pointer can't change". It means "you can't write through this pointer". ++s1 is fine. s1 = '\0', however, is not. And since you don't want to do that anyway, I suggest that you use const char - in which case pS1 and pS2 become redundant; you might as well use s1 and s2 directly. (Remember that C is pass-by-value, so ++s1 only changes your local copy of the pointer, not the value of the pointer passed by the caller.) <snip> -- Richard Heathfield "Usenet is a strange place" - dmr 29/7/1999 http://www.cpax.org.uk email: rjh at the above domain, - www. - Hide quoted text - Peithon <Peit@googlemail.com> writes: > Hi, > This is a very simple question but I couldn't find it in your FAQ. > I'm using VC++ and compiling a C program, using the /TC flag. > I've got a function for comparing two strings > int strspcmp(const char * s1, const char * s2) > { > char* pS1 = s1; > char* pS2 = s2; > ... > ... > } > As you can see, I assign the two const ptrs to some non-const ptrs to > do some ptr arithmetic, > like incrementing the ptrs to compare the strings. > But I'm getting warning C4090 complaining about the assignment. > I thought I was doing the standard thing for ptr arithmetic. What am I > doing wrong? > How do I get rid of the warning? For a start you post the warning you are getting :-; Peithon <Peit@googlemail.com> writes: > Hi, > This is a very simple question but I couldn't find it in your FAQ. > I'm using VC++ and compiling a C program, using the /TC flag. > I've got a function for comparing two strings > int strspcmp(const char * s1, const char * s2) You are invading the implementation's name space. > { > char* pS1 = s1; > char* pS2 = s2; > ... > ... > } > As you can see, I assign the two const ptrs to some non-const ptrs The pointers are not const. They point to const char. Subtle but important distinction. > to do some ptr arithmetic, Odd. You can do arithmetic on const char * as well and you would not get the warnings. > like incrementing the ptrs to compare the strings. > But I'm getting warning C4090 complaining about the assignment. As you must. It requires a diagnostic. > I thought I was doing the standard thing for ptr arithmetic. What am I > doing wrong? > How do I get rid of the warning? Just to the pointer arithmetic on the const char s1 and s2. -- Ben. On 1 Jun, 18:47, Ben Bacarisse <ben.use@bsb.me.uk> wrote: - Hide quoted text - > Peithon <Peit@googlemail.com> writes: > > Hi, > > This is a very simple question but I couldn't find it in your FAQ. > > I'm using VC++ and compiling a C program, using the /TC flag. > > I've got a function for comparing two strings > > int strspcmp(const char s1, const char * s2) > You are invading the implementation's name space. > > { > > char* pS1 = s1; > > char* pS2 = s2; > > ... > > ... > > } > > As you can see, I assign the two const ptrs to some non-const ptrs > The pointers are not const. They point to const char. Subtle but > important distinction. > > to do some ptr arithmetic, > Odd. You can do arithmetic on const char * as well and you would not > get the warnings. > > like incrementing the ptrs to compare the strings. > > But I'm getting warning C4090 complaining about the assignment. > As you must. It requires a diagnostic. > > I thought I was doing the standard thing for ptr arithmetic. What am I > > doing wrong? > > How do I get rid of the warning? > Just to the pointer arithmetic on the const char s1 and s2. > -- > Ben. Thanks, I didn't see the distinction between const char and char * const. Much appreciated. - Hide quoted text - Peithon wrote: > Hi, > This is a very simple question but I couldn't find it in your FAQ. > I'm using VC++ and compiling a C program, using the /TC flag. > I've got a function for comparing two strings > int strspcmp(const char * s1, const char * s2) > { > char* pS1 = s1; > char* pS2 = s2; > ... > ... > } > As you can see, I assign the two const ptrs to some non-const ptrs to > do some ptr arithmetic, > like incrementing the ptrs to compare the strings. > But I'm getting warning C4090 complaining about the assignment. > I thought I was doing the standard thing for ptr arithmetic. What am I > doing wrong? > How do I get rid of the warning? typing const char a; means you have a pointer to a char and are not allowed to modify what it points to (but you are allowed to modify the pointer). that means char b; a="hello"; a++; b=a; is valid code, but not: a=&b; a=' '; / this line is not valid / since this is valid: char c; c=&b; c=' '; but modifying something pointed to by a const pointer is not is this also not valid: a=&b; c=a; / gives the warning you mentioned / c=' '; /* this statement is still valid / - Hide quoted text - Peithon <Peit@googlemail.com> writes: > This is a very simple question but I couldn't find it in your FAQ. > I'm using VC++ and compiling a C program, using the /TC flag. > I've got a function for comparing two strings > int strspcmp(const char s1, const char * s2) > { > char* pS1 = s1; > char* pS2 = s2; > ... > ... > } > As you can see, I assign the two const ptrs to some non-const ptrs to > do some ptr arithmetic, > like incrementing the ptrs to compare the strings. > But I'm getting warning C4090 complaining about the assignment. > I thought I was doing the standard thing for ptr arithmetic. What am I > doing wrong? > How do I get rid of the warning? This has already been ably answered, but perhaps a simpler example might make the point clearer. I think your point of confusion is that "const char" means "pointer to const char", not "const pointer to char", but I'll explore other issues as well. int main(void) { const char ptr = "hello"; char copy_of_ptr = ptr; ptr = 'H'; return 0; } The declaration "const char ptr" doesn't make ptr const; it makes what it points to const. Here I've initialized ptr to point to a string literal, i.e., to an array of 6 characters 'h', 'e', 'l', 'l', 'o', '\0'. This array could be stored in read-only memory; attempting to modify it could cause the program crash (it's undefined behavior). The declaration of copy_of_ptr doesn't use the "const" keyword, which means I'm allowed to modify either the pointer object itself or* what it points to. By "allowed to", I mean that the compiler won't complain if I try to do it; it will go ahead and generate code to do what I asked. It's up to me to make sure that I actually can modify whatever I'm trying to modify. If I lie to the compiler, it may or may not complain. But in this case, I tried to lie to the compiler and it caught me at it. I attempted to initialize a char* to the value of ptr. If I could do this, it would allow me to (attempt to) modify whatever copy_of_ptr points to. And that's exactly why the language forbids assigning the value of ptr to copy_of_ptr (it's a constraint violation, requiring a diagnostic). My error is caught during compilation, when I attempt to copy the pointer, which is much better than not catching it until run time, when I actually try to modify the string. If the language, and therefore the compiler, allowed me to copy the value of ptr to copy_of_ptr, then it would also allow me to attempt to modify the string that it points to (remember that it might be in read-only memory), all without telling me that I've done something dangerous. Fortunately, the rules of the language are such that I can't get away with this particular error quite so easily. (It is overly lax about string literals, but that's another story.) If you want a pointer to const char, declare it as: const char ptr; If you want a const pointer to char (you can't modify the pointer, but you can modify what it points to): char const ptr; And if you want both the pointer and what it points to be read-only: const char const p; The placement of the "const" keyword is tricky. If you can find a program called "cdecl", it can help you with this kind of thing; I used it to confirm these declarations. % cdecl Type `help' or `?' for help cdecl> declare p as const pointer to char char const p cdecl> declare p as const pointer to const char const char * const p cdecl> explain const char p declare p as pointer to const char cdecl> One thing to watch out for: cdecl treats "ptr" as a synonym for "pointer", so if you tell it "declare ptr as pointer to const char" it will report a syntax error. -- Keith Thompson (The_Other_Keith) k@mib.org <http://www.ghoti.net/~kst> San Diego Supercomputer Center <> <http://users.sdsc.edu/~kst> "We must do something. This is something. Therefore, we must do this." -- Antony Jay and Jonathan Lynn, "Yes Minister" Keith Thompson wrote: ... snip ... > One thing to watch out for: cdecl treats "ptr" as a synonym for > "pointer", so if you tell it "declare ptr as pointer to const > char" it will report a syntax error. [OT] A devilish concoction. Not mentioned in the help material. And there are others of this type. -- <http://www.cs.auckland.ac.nz/~pgut001/pubs/vista_cost.txt> <http://www.securityfocus.com/columnists/423> <http://www.aaxnet.com/editor/edit043.html> <http://kadaitcha.cx/vista/dogsbreakfast/index.html> cbfalconer at maineline dot net -- Posted via a free Usenet account from http://www.teranews.com - Hide quoted text - Peithon wrote: > This is a very simple question but I couldn't find it in your FAQ. > I'm using VC++ and compiling a C program, using the /TC flag. > I've got a function for comparing two strings > int strspcmp(const char * s1, const char * s2) > { > char* pS1 = s1; > char* pS2 = s2; > ... > } > As you can see, I assign the two const ptrs to some non-const ptrs > to do some ptr arithmetic, like incrementing the ptrs to compare > the strings. But I'm getting warning C4090 complaining about the > assignment. I thought I was doing the standard thing for ptr > arithmetic. What am I doing wrong? How do I get rid of the warning? Declare the local variables as const: const char pS1 = s1; etc. You have declared the input as pointers to const, which are therefore not modifiable. -- <http://www.cs.auckland.ac.nz/~pgut001/pubs/vista_cost.txt> <http://www.securityfocus.com/columnists/423> <http://www.aaxnet.com/editor/edit043.html> <http://kadaitcha.cx/vista/dogsbreakfast/index.html> cbfalconer at maineline dot net -- Posted via a free Usenet account from http://www.teranews.com - Hide quoted text - Richard wrote: > Peithon <Peit@googlemail.com> writes: >> This is a very simple question but I couldn't find it in your >> FAQ. I'm using VC++ and compiling a C program, using the /TC >> flag. I've got a function for comparing two strings >> int strspcmp(const char s1, const char * s2) >> { >> char* pS1 = s1; >> char* pS2 = s2; >> ... >> } >> As you can see, I assign the two const ptrs to some non-const >> ptrs to do some ptr arithmetic, like incrementing the ptrs to >> compare the strings. But I'm getting warning C4090 complaining >> about the assignment. I thought I was doing the standard thing >> for ptr arithmetic. What am I doing wrong? How do I get rid of >> the warning? > For a start you post the warning you are getting :-; You mean you can't see the problem? -- <http://www.cs.auckland.ac.nz/~pgut001/pubs/vista_cost.txt> <http://www.securityfocus.com/columnists/423> <http://www.aaxnet.com/editor/edit043.html> <http://kadaitcha.cx/vista/dogsbreakfast/index.html> cbfalconer at maineline dot net -- Posted via a free Usenet account from http://www.teranews.com