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What is the difference between char array[] and char *p in C

1 Answer

0 votes
#include <stdio.h> 

int main(void) 
    char arr[] = "c c++ java php";
	char *p = "c c++ java php";
    printf("1. %s\n", arr);
	printf("2. %s\n\n", p);
	printf("3. %d %d\n", (int)sizeof(arr), (int)sizeof(p));
	printf("4. %p %p\n", arr, p);
	printf("5. %p %p\n", &arr, &p);
	printf("6. %c %c %c\n\n", arr[0], p[0], *(p + 0));
	arr[0] = 'x';
	//p[0] = 'x'; Error
    printf("7. %s\n", arr);
	printf("8. %s\n\n", p);
	printf("9. %c %c %c\n\n", arr[13], p[13], *(p + 13)); // last char
	printf("10. %c %c %c\n\n", arr[14], p[14], *(p + 14)); // arr[14]=null (end string) p[14]=?
	printf("11. %c %c %c\n\n", arr[15], p[15], *(p + 15)); // arr[15]=? (out of index) p[15]=?
	//arr++; Error
    printf("12. %s\n", arr);
	printf("13. %s\n\n", p);

    return 0; 
1. c c++ java php
2. c c++ java php

3. 15 8
4. 000000000062FE10 0000000000404030
5. 000000000062FE10 000000000062FE08
6. c c c

7. x c++ java php
8. c c++ java php

9. p p p


11.   1 1

12. x c++ java php
13.  c++ jaava php


answered Aug 11 by avibootz
edited Aug 11 by avibootz