I thought str1 is the destination and will be modified, i.e. appended with str2. And using two const char * I got warnings as:
Code:
prog009c2.c:73:2: warning: passing argument 1 of ‘strmerg' discards ‘const' qualifier from pointer target type [enabled by default]
prog009c2.c:21:7: note: expected ‘char *' but argument is of type ‘const char *'
Is it? I didn't think it changed, but that may be my mistake. In any case, you have to think about the code I give you, not blindly use it -- especially not make blind changes just to "fix compiler errors". Can you tell me why I said to make these variables const? And how I told you to deal with those warnings?
As for question 2 -- you're back at square one with pointers again. What, exactly, have you assumed that str3 = ... is going to do?
Last edited by Corona688; 04-02-2014 at 01:44 PM..
My understanding of "why they are const" is to avoid modification of the parameters by accident. For my strmerg( char *dest, char *src) function, the original design is dest and src are inter-changable, i.e. src can be appended to dest and vice verse.
I did not know the const char * restriction at the beginning to avoid editing the parameters.
True, at this moment I am trying to cross the stage to "make blind changes ...to fix the compiler error". When you ask me "why I said make these variables const" I thought I understood your saying, but not really. ------Should I give up with C now? Thank you!
I am dumping raw memory contents to show you what happens when you declare a variable on the stack, and allocate memory with malloc.
For the purposes of this, you can ignore the contents of the 'printpage' and 'spew' functions, they're convenience functions I made to dump memory and are not related otherwise.
Remember that memory can be considered to be one gigantic array of bytes, from address 00000000 all the way up to ffffffff (on a 32-bit machine). Pointers are just array indexes inside this array.
So we allocate four blocks of 16 bytes, set their first element to something, and dump memory to see where they ended up:
...but what about mem1 itself? Something, somewhere, has to remember that address of 0x85ef008, yes? And it does, on the "stack", which is a big block of memory which the processor uses as temporary space.
There's actually quite a lot of stuff on the stack. It's not just us that's using it. Every time you call a function, it uses stack to pass arguments, create local variables, and remember where to return. It gets used so much, in fact, that you can't trust that a local variable doesn't contain previously-used garbage values unless you set it to anything else yourself.
It's there all right. Slightly out-of-order, but it's there. That's the order an x86 processor handles all numbers, nothing weird. Humans are weird in wanting it highest-to-lowest digit order instead of something easily mechanically processable, which is why we have printf to handle that job for us.
Now, we want to put a different string into mem1. What will the statement 'mem1=mem3' do?
If mem1 is a pointer to a string, mem1=... does not modify the string. It alters the pointer.
Moderator's Comments:
pointer = cannot, will not, absolutely not, at all, ever, even for the sake of argument ever do anything else except alter the pointer.
Also: free() requires the exact same pointer that malloc() gave you. If you give it a different pointer, even slightly, it will crash.
If you give it the same pointer twice, it will crash. (Which means, if we did free() on all our pointers, our program would crash right now.)
If you write beyond the end of the memory you allocated, it will probably crash. (Because, as you can see in the dump, if you write beyond that you're stomping on top of something else).
Last edited by Corona688; 04-02-2014 at 03:15 PM..
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We've had this conversation four or five times. You have a hard time telling when you are modifying the pointer instead of its contents. But the compiler can tell you that easily, so I have a suggestion:
Whenever you do char *mem=malloc(300); ...do this instead: char * const mem=malloc(300); This will make the mistake you keep repeating a compiler error -- "assignment of read-only variable". (It of course sets it, once, when you declare it. But thereafter it's considered 'fixed'.)
You are still free to modify its contents, like with strcpy(mem, originalstring) or mem[5]='Q' or *(mem+5)=37 or any other way you please. But if you try to alter the pointer that's an error. You can understand "assignment of read-only value" to mean "whoops, I mixed up a pointer's value and its contents".
If you find yourself needing to do complicated circumlocutions to get around 'assignment of read only variable', you can be fairly sure you've taken a wrong turn somewhere.
Last edited by Corona688; 04-02-2014 at 03:34 PM..
These 2 Users Gave Thanks to Corona688 For This Post:
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04-02-2014
