指向常量的指针与打印时指向非常量的指针的行为不同

Question

I have the following code:

int main(int argc, char **argv)
{
printf("%s\n",*argv);
int test = 5;
char* p;
*pint = test;
p = "banana";
printf("%s\n",p);
printf("%d\n",*pint);
}

Why is it that I have to write p="banana" and not *p="banana" but for an integer, it needs to be *pint , otherwise it will only print the address of the integer? Shouldn't p print the address of "banana" ?

Answer 1

1. You are comparing array and integer variable behavior !

2. p = "banana";

You are assigning base address of string "banana" to pointer p.

And the printf function prototype is

int printf( const char *restrict format, ... );

printf("%s\n",p);

Above statement implies that you are passing pointer p as a argument to a function printf which holds address of string "banana"

Answer 2

You are printing with %s . This prints a C string taking an input as the address of the first byte of the string.

If you print it with %p , you will get the address.

printf("%p\n",p);

Answer 3

A C-style string is an array of characters terminated by '\\0' . So when you assign a string to it, this is what it looks like;

char p[] = {'b', 'a', 'n', 'a', 'n', 'a', '\\0'};

So when say, you print out p using format specifier %s , it will continue to print the remaining characters till it reaches the null terminating character.

Doing this printf("%c", *p) will only print the first character.

Using an integer, if you do this;

int p[] = {1,2,3,4,5};

And print it out;

print("%d", *p);

You only get the first integer in the array.

Note; each format specifier has its kind of value it accepts. That is why they're called format specifiers

PS:

I've edited my answer based on user2079303 and alk's comments!