C programming: printing a null-terminated string using a character pointer

Question

I am just getting used to pointers in C. I was thinking of using %s to print the first character of a character array.

In the following code, why does %s not work, but %c works? If *my_string is dereferencing the first value of the character array, wouldn't %s try to print all the characters in memory starting at that address and continue until a null-zero character is reached?

Also, how is printf("%s", my_string) working, but printf("%s" *my_string) is not?

char* my_string = "The original value.";
printf("Value of my_string: %s\n", my_string);
my_string = "The new value.";
printf("Value of my_string: %s\n", *my_string); // Error
printf("Value of my_string: %c\n", *my_string); // Prints only the first letter 'T', but does not have any errors

%s expects a null terminated string, *my_string is not a pointer, it is a char value, and %s tries to use that value as a pointer, which is obviously wrong. — Moshe Gottlieb
– Moshe Gottlieb, Commented Aug 20, 2020 at 6:13
" If *my_string is dereferencing the first value" It is. Meaning it goes to the address and fetch the contents there. "wouldn't %s try to print all the characters in memory starting at that address" But my_string is the address, *my_string is the contents of that address. %s expects the former, not the latter. — Lundin
– Lundin, Commented Aug 20, 2020 at 10:24

Cornholio · Accepted Answer · 2020-08-20 06:13:52Z

1

The asterisk * de-references the pointer to the type that it is pointing to, so when my_string is a pointer to a character, *my_string is a character. Your C compiler sees your format string containing a %s and a parameter of *my_string which is a character and shows you an error.

To output the whole string, print it as in your first call to printf, to print only a single character do as you did in your last call to printf.

answered Aug 20, 2020 at 6:13

Cornholio

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Zongru Zhan · Accepted Answer · 2020-08-20 06:26:22Z

1

why does %s not work, but %c works?

It is the rule by which how the printf function works. For "%s" it expects a pointer to a null ending string, and for "%c" it expects a single character.

If *my_string is dereferencing the first value of the character array, wouldn't %s try to print all the characters in memory starting at that address and continue until a null-zero character is reached?

Yes. So you must be careful in case it is a string too long and these characters are printable before the null ending.

Also, how is printf("%s", my_string) working, but printf("%s" *my_string) is not?

For "%s", it expects a pointer. "*my_string" is a character.

answered Aug 20, 2020 at 6:26

Zongru Zhan

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1 Comment

chux Over a year ago

Good first answer.

Marko Borković · Accepted Answer · 2020-08-28 16:03:05Z

1

To understand this we need to go into how the string "The original value." is stored in the memory and what the char* my_string points to.

We can expect a memory structure like this

0xDEADBEEF    ASCII T
0xDEADBEEF+1  ASCII h
0xDEADBEEF+2  ASCII e
0xDEADBEEF+3  ASCII <space>
...
0xDEADBEEF+16 ASCII u
0xDEADBEEF+17 ASCII e
0xDEADBEEF+18 ASCII .
0xDEADBEEF+19       0

So when you have a char* my_string it points to the first character or in this case the address 0xDEADBEEF. So when you use %s in a printf it will loop through all the characters starting from the first one you pointed to until it reaches \0. But *my_string just reads the first character on that address. For that reason we need to give it a pointer.

edited Aug 28, 2020 at 16:03

answered Aug 20, 2020 at 6:22

Marko Borković

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