printf以不同的方式表现……并采用地址和值……字符串的地址和指针的地址，仅在解引用之后

Question

printf is taking address of string and not taking after dereferencing, whereas in case of pointer it is required to dereference.

#include <stdio.h>
#include<stdio.h>

int main()
{
    char str[100];
    int i;
    int j=0;
    int *p;

    p=&j;
    scanf("%s",str);
    for(i=0;str[i];i++)
    {
        if((str[i]>='A') && (str[i]<='Z'))
        {
            str[i]=str[i]+('a'-'A');
        }
        else
        {
            str[i]=str[i]-('a'-'A');
        }
    }
    printf("%s",str);   //it should have been printf("%s",*str); here we are  passing address
    printf("%d\n",j);
    printf("%d",*p);    //here we are passing evact value;

    return 0;
}

when used with * it crashes and if only str is used it works fine...

Answer 1

The %s format specifier to printf is used to print a string and expects a char * argument which points to the first element of a null-terminated character array. The %d format specifier is used to print an integer in decimal format and expects an int .

Since str is an array, when used in an expression it decays into a pointer to its first element. So str in an expression has type char * , which matches what %s expects.

*str is not valid for %s because that has type char and has a value of the first character in an array. Using the wrong format specifier for a given argument to printf invokes undefined behavior .

*p is valid for %d because p has type int * , meaning that *p has type int , which matches what %d expects.