strdup and malloc

[ueg1990]

hello,

i was looking at the code online for strdup and wanted to ask a question regarding its implementation w.r.t malloc()

strdup
 merely copies the string given by its argument into a safe place, obtained by a call on 
malloc
:
   char *strdup(char *s)   /* make a duplicate of s */   {       char *p;       p = (char *) malloc(strlen(s)+1); /* +1 for '\0' */       if (p != NULL)           strcpy(p, s);       return p;   }

I wanted to know why is there a need to call malloc on the variable p? wont this function still work without the malloc?

hope to hear from you soon.

[ueg1990]

   char *strdup(char *s)   /* make a duplicate of s */   {       char *p;       p = (char *) malloc(strlen(s)+1); /* +1 for '\0' */       if (p != NULL)           strcpy(p, s);       return p;   }

[ledow]

strcpy does not allocate memory, it just copies what it's told to, where it's told to copy it. Malloc actually allocates memory so that strcpy has somewhere to copy TO.

[ueg1990]

So, if we do not malloc, it will not copy from s to p?

[stahta01]

So, if we do not malloc, it will not copy from s to p?

If you DO NOT do malloc inside the strdup function then there is nowhere legal/valid/safe to copy the string.

Tim S.

[ueg1990]

oh ok...thnx!!

btw nice quote!!!

[hk_mp5kpdw]

Without the malloc call, the strcpy will be copying from source string s to an uninitialized pointer p. Being uninitialized, p could be pointing anywhere, most likely somewhere that will cause a segfault if you attempt to write to it. A segfault would be a good option as you'd know you were doing something wrong immediately. A bad option would be the strcpy would silently trash some other critical data without you knowing anything was amiss leaving your program to merrily go on about its business without apparent fault.