Why can't I printf the pointer at %s?

I have a question when I see 's answer that I don't know how to behave when I refer to charstr[] with a pointer.

Why shouldn't I write printf("str:%s\n",p);?

*p calls the character code for the address in memory.So I thought p would be fine, but the debug will not work even if there is no error.If I were to use *p, how would I change the printf operator?

Also, why is &(*p) good?Is it because the operator can only receive the character code from the address?In printf("*p:%s\n", &(*p)), I tried printf("*p:%s\n", &p), but it didn't work.Why is &(*p) of printf("*p:%s\n", &(*p)) OK?Or is it possible to change the operator %s of printf("*p:%s\n", &p) with &p?

#include<stdio.h>
int main(void){
    /*str[ ] ensures that there are 5 consecutive characters of addresses in memory and the contents are abcde.*/
    char str [ ] = "abcde";

    /* Let's look at the first memory address saved in str one by one.
    /* The loop doesn't end until NULL comes out. */
    /* From the first character to the terminating NULL in this loop.Displays the second character to the terminating NULL in order */
    for (char*p=str;*p;p++)
    {
        /* The pointer*p provides direct manipulation of the contents of the address on the memory reserved by str[].*/
        /* The pointer allows you to directly manipulate the memory addresses secured by str[].*/
            /* Take the value of the memory address indicated by *p and add 1 to it */
            /* It is substituted for the address indicated by *p.The contents of the memory saved by str are rewritten.*/
        (*p) = (*p) + 1;
        /* Displays the address indicated by *p up to the end of NULL.*/
        printf("*p:%s\n", &(*p));
        /* Also displays the contents of str.*/
        printf("str:%s\n", str);
    }
}

c pointer

2022-09-30 16:15

2 Answers

Why shouldn't I write printf("str:%s\n",p);

What made you think so?I think there is no problem with the code.
printf("*p:%s\n", &(*p)) is good and printf("str:%s\n",p); is not bad.
If anything, printf("*p:%s\n", &(*p)) is more problematic.
&(*p)) may be intended to indicate that it is an address, but it is not usually written in this way.

In char*p;, &(*p) and p are the same.By the way, &(*(&(*p))) is the same as p.
&Area means the address of the area, and *Address is the content of the area that the address points to.

If p contains a char-type area address, contains zero or more character codes in that area, and is zero-terminated, printf("str:%s\n", &(*p)) works fine as well as printf("str:%s\n",p);.

2022-09-30 16:15

I don't remember saying no, but if you want to take it and solve it yourself, check it out yourself until you realize there's no problem.

Is there an environment where I can check it?

The printf is also detailed in Wikipedia.
If you feel like it, you can check the definition files of various operating systems.
I think you can check the definition of the header.

2022-09-30 16:15

If you have any answers or tips

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