CODEWITHSUNDEEP
cstringpointerscharprintf
Tilo
Tilo

Reputation: 45

C - format specifies type 'int' but the argument has type 'char *' [-Wformat]

I have defined a vector of pointers with weekdays. Later on, I want to sort the days lexicographically.

I print all the weekdays by using printf("%s", *(wochentag + i)); which works. I want the output of each first letter of each weekday. Later, I want to also access the second, third, ... letter of each word.

By using 

printf("%c", *(wochentag + i));`

I get the following warning:

format specifies type 'int' but the argument has type 'char *' [-Wformat].

Here's my code:

int main()
{
   static char *wochentag[] = {"Montag", "Dienstag", "Mittwoch", "Donnerstag", 
                        "Freitag", "Samstag", "Sonntag"};
    printf("%lu\n", sizeof(wochentag));
    printf("%lu\n", sizeof(char));
    for (int i = 0; i < 7; i++)
    {
        printf("%s\n", *(wochentag + i));
    }
    for (int i = 0; i < 7; i++)
    {
        printf("%c\n", *(wochentag + i));
    }
}

What am I doing wrong?

Upvotes: 1

Views: 1949

Answers (3)

anastaciu
anastaciu

Reputation: 23832

In printf("%c\n", *(wochentag + i)), *(wochentag + i) is a string, more precisely a dereferenced pointer to a string literal (null-terminated read-only byte array), as you can see in your code, you need "%s" format specifier.

To print the first char in each string you would need an extra dereference: 

printf("%c\n", **(wochentag + i));

Or for a more reader-friendly notation

printf("%c\n", wochentag[i][0]);

The warning indicating that "%c" expects an int is because it is indeed expecting the int code for the specific character.

A practical example:

int x = 65;  //ASCII code for the letter 'A'
printf("%c\n", x);

This will output A.

Upvotes: 1

Yousaf Raza
Yousaf Raza

Reputation: 793

Your array data type char*[] is a type of double pointer so when you print using %s, it knows that it has address of some array of chars but when you print using only '%c' in printf you have to give the actual character but not the address....so you have to de-reference it twice

Like below

printf("%c", **(wochentag + i));

Hope this answers your problem....peace

Upvotes: 0

Sourav Ghosh
Sourav Ghosh

Reputation: 134416

The error message is self-explanatory, there is a data type mismatch.

Another syntax of writing *(wochentag + i) is (wochentag[i]), which is of type char *.

Now, passing that to %c is wrong, as this conversion specification expects a char type.

If you're interested in only the first char, you have to index into that array, like (wochentag[i][0]). Something like

 printf("%c\n", wochentag[i][0]);

should work.

Upvotes: 2

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