CODEWITHSUNDEEP
cpointersgccsyntaxnull
backus
backus

Reputation: 4316

Checking pointer of *char[] is NULL without upsetting gcc

I have char *words[] that holds strings of varying length.

Let words contain 5 dynamically allocated strings: ["zero","one","two","three","four"].

I free one of the strings and set its pointer to NULL like so:

free(words[1]);
*(words[1]) = NULL;

Later on I loop through the words array and check if each item is NULL like so:

if(*words[i] == NULL)

And when I compile gcc says warning: comparison between pointer and integer. A bit of googling has shown me that I can silence this warning by changing the check to:

if(*words[i] == 0) or if(*words[i])

I know its nitpicky, but these checks just aren't as clear as it would be if it used NULL. Is there a way to compare using NULL without altering gcc flags or using a #define-esque solution?

As a reference, my compile command is:

gcc -m32 -g -O0 -std=gnu99 -Wall -Wfloat-equal -Wtype-limits -Wpointer-arith -Wlogical-op -fno-diagnostics-show-option  -m32  words.c   -o words

Upvotes: 1

Views: 517

Answers (1)

John Colanduoni
John Colanduoni

Reputation: 1626

The code isn't doing what you think it is. *words[i] gives you the first character in the ith string (i.e. a char), not the ith string itself (i.e. a char*). What you want is just words[i]. Remember that in C char* words[] is basically char** words.

The warning comes from the fact that C considers char to be an integer. In this case, it is a good thing the compiler warned you!

Upvotes: 7

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