Print char pointer char array with printf?

Question

I'm trying to print the whole word instead of just the first letter. I could do it with a loop but I figured there was a better way. I was searching around and saw answers that they changed %S to %c but I'm already using %c since it's a character array.

char* words[] = {"my", "word", "list"};
printf("The word: %c",*words[2]);

Results:
The word: l

Mel Nicholson · Accepted Answer · 2012-11-08 20:54:43Z

9

The issue is that you dereferenced twice. The [2] in *words[2] dereferences from words[] to "list" then the * dereferences a second time from "list" to 'l' Remove the * and voila.

char* words[] = {"my", "word", "list"};
printf("The word: %s", words[2]);

answered Nov 8, 2012 at 20:54

Mel Nicholson

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krizzo Over a year ago

Ah I see, I was wondering about the dereferencing but I was getting random characters if I didn't before when I was using %S I might have read the lowercase as an upper case. Thank you that solved it.

Sergey Kalinichenko · Accepted Answer · 2012-11-08 20:52:29Z

4

You need to use %s, a format used specifically for null-terminated arrays of characters (i.e. C strings). You do not dereference the array's element when you pass it to printf, like this:

printf("The word: %s\n", words[2]);

answered Nov 8, 2012 at 20:52

Sergey Kalinichenko

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krizzo Over a year ago

I must have read the code wrong when I did my search I did have %S but as you can see I used uppercase. I was getting random characters which I figured it was displaying from the address which is why I double derefferenced it. Removed that and changed it to s now everything works. Thanks.

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