A trivial question?

[Protonman]

Hi,

I'm asking this question out of curiosity rather than neccessity:

E.g:

------
//Vars

char foo[20] = "abcd";
char *bar = "ghjk";


a) scanf("%s",bar); /*doesn't work; compiles but crashes (using g++) */

b) strcat("qwerty",foo); /* ditto, although foo shouldn't change */

c) strcat(bar,foo); /*" "; " " "*/
strcpy(...); /* ... */

d) strcat(foo,bar); /*works, no problem*/
strcpy(...); /* ... */

-----

The same "problem" arises when using other string functions such as strcpy etc. Also, why can't you "scanf" into a char *, only into a char * constant (array name)?

Is this because "bar" is a pointer and "foo" a constant pointer?

I'm just very curious to know why and can't seem to figure it out.

Many thanks in advance

Protonman

 

[Salem]

All your char * point to string constants, which implementations are free to make read-only.
This means that if you try, you get a segmentation fault as the OS steps in and kills your program.

If you point at memory you can modify, say an array, then all is well.

  char foo[20] = "abcd";
  char *bar = foo;

You can scanf into bar, and it will just write over whatever foo is initialised to.

--

 

[maur3r]

The solution to ""modify"" the content of * is to allocate memory for it. (functions malloc, calloc, realloc). Personaly I prefer this solution because you don't have to declare size of the array before compilation and can increase it dynamically.

char *bar;
char *tmpbar;
bar=malloc(10/*any number*/*sizeof(char));/*equal to bar[10]*/
bar[0]='c';/*now it's legal*/
tmpbar=realloc(bar,(10+10)*sizeof(char));/*if you wan't bigger size*/
/*tmpbar is used because when you run out of memory realloc returns NULL and all data is lost (memory leak)*/
if(tmpbar)
bar=tmpbar;
free(bar); /*dispose the pointer at the end*/

Martin