I have a variable (Mydict) which is struct type, which includes a pointer (dict0) to an array of chars (5 of them). now I want to assign a word to that array and can't figure out how to do it.
I can't change the lines in the struct definition itself so I need to know how to write the last line correctly (the one that's supposed to assign "olga" to the array.
Code:int main(){ struct dict{ int length; char (*dict0)[5]; }; struct dict Mydict; Mydict.dict0="Olga"; return 0; }
strcpy is the answer.
In C we have to use the string.h library for string handling.
Also the struct does not seem ok.
First of all, when we want to access a pointer into a struct we use the operator -> , instead of the dot.
Secondly what you need, is just an array of characters.
Try naming your variables in a more meaningful wayCode:char dict[5];
Code - functions and small libraries I use
It’s 2014 and I still use printf() for debugging.
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> char (*dict0)[5];
This is a pointer (just one) to an array of 5 chars.
Contrast with char *dict0[5] which is an array of 5 pointers to char
You would need
Code:Mydict.dict0 = malloc( sizeof(*Mydict.dict0) ); strcpy(Mydict.dict0,"test"); // copy at most 4 chars
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I don't quite understand why I need malloc. don't I already assign the memory in the struct definition when I define the dict0 points to 5 characters long array?
thanx Salem. I think you just solved my problem. although I had to add '*' before Mydict in the strcpy function.
If you don't want to use malloc you need a buffer somewhere that you can point to.Originally Posted by littlerunaway
If you don't want to use the temporary variable you can also use the typecastCode:static char buffer[5] = {0}; int main(void){ struct dict{ int length; char (*dict0)[5]; }; struct dict Mydict; { char (*ptr)[5] = &buffer; Mydict.dict0 = ptr; } strcpy(*Mydict.dict0, "test"); printf("*Mydict.dict0: %s\n", *Mydict.dict0); return 0; }
Mydict.dict0 = (char(*)[5])&buffer;
But this looks quite ugly to me. What's wrong with just making your member a char** ?
I don't see why you cannot just write:
since buffer is a char[5], hence &buffer is already a char(*)[5].Code:Mydict.dict0 = &buffer;
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That's correct. However in the general case you probably would declare a larger buffer and want to allocate from it
Then you would need the typecast, for example, to allocate 5 bytes from offset 100:Code:static char buffer[10000] = {0};
That's why I would say to make it char ** to simplify the notation. Or consider typedef if you really need the [5] for compiler type safety.Code:Mydict.dict0 = (char(*)[5])&(buffer[100]);
Thank you all! It is usefull for me to fix my programming errors.
two accounts? Not good.
Code - functions and small libraries I use
It’s 2014 and I still use printf() for debugging.
"Programs must be written for people to read, and only incidentally for machines to execute. " —Harold Abelson
