Printing a Char *

Question

I apologize in advance for the dumb question!

Here is my struct def:

struct vcard { char *cnet; char *email; char *fname; char *lname; char *tel; };

I am trying to print a representation of this struct with the function vcard_show(vcard *c), but the compiler is throwing back an warning:

void vcard_show(struct vcard *c) { printf("First Name: %c\n", c->fname); printf("Last Name: %c\n", c->lname); printf("CNet ID: %c\n", c->cnet); printf("Email: %c\n", c->email); printf("Phone Number: %c\n", c->tel); }

When compiled: "warning: format ‘%c’ expects type ‘int’, but argument 2 has type ‘char *’"

Isn't %c the symbol for char*?

Jonathon Simister · Accepted Answer · 2013-03-08 00:20:44Z

You want to use %s, which is for strings (char*). %c is for single characters (char).

An asterisk * after a type makes it a pointer to type. So char* is actually a pointer to a character. In C, strings are passed-by-reference by passing the pointer to the first character of the string. The end of the string is determined by setting the byte after the last character of the string to NULL (0).

DanZimm · Accepted Answer · 2013-03-08 00:20:35Z

The property type encoding for a char * is %s. %c is for a char (not the pointer just a single char)

msc · Accepted Answer · 2017-05-25 10:15:18Z

Unless you have some typedef you are not telling us about, you should probably declare vcard_show() like this:

void vcard_show(struct vcard *c)

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