C - Copy string into a linked list variable

Question

#define STRLEN 65

/*Create linked list */
struct node {
   char str[STRLEN];
   struct node *next;
};

newInput = malloc(sizeof(struct node));
strcpy(newStr, newInput->str);

I left the other parts of the code, but it doesnt seem to be copying the string into newInput->str.

The string accepted is only 64 bytes.

It's just blank when I print it out after the copy. Any clue why?

Mark Elliot · Accepted Answer · 2010-10-05 02:35:59Z

4

You have the arguments to strcpy reversed, the first argument is the destination, the second argument is the source. Try:

strcpy(newInput->str, newStr);

answered Oct 5, 2010 at 2:35

Mark Elliot

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2 Comments

aschepler Over a year ago

Or even better, strncpy(newInput->str, newStr, STRLEN);

R.. GitHub STOP HELPING ICE Over a year ago

strncpy is the wrong function to use here. It zero-fills the remainder of the size, which is a waste of time, and if newStr were longer than the destination size, it would not get null-terminated. Never use strncpy unless you really know what it was designed for.

vanza · Accepted Answer · 2010-10-05 02:36:37Z

0

You have the arguments to strcpy reversed.

answered Oct 5, 2010 at 2:36

vanza

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Comments

codaddict · Accepted Answer · 2010-10-05 02:36:49Z

0

The first parameter of strcpy is the destination and the 2nd param is the source.

So

strcpy(newStr, newInput->str);

should be

strcpy(newInput->str,newStr);

answered Oct 5, 2010 at 2:36

codaddict

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Comments

JoshD · Accepted Answer · 2010-10-05 02:36:50Z

0

Your destination and source are backwards:

strcpy(newStr, newInput->str);

should be

strcpy(newInput->str, newStr);

answered Oct 5, 2010 at 2:36

JoshD

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