Linux C language: strings and arrays

1. The arrangement of the memory array affirmed

The variable size of the array can not be used;

#include <stdio.h>
int main()
{
	int a=3;
	int b=2;
	int array[3];
	array[0]=1;
	array[1]=10;
	array[2]=100;
	int *p=&a;
	int i;
	for(i=0;i<6;i++)
	{
		printf("*p=%d\n",*p);
		p++;
	}
	printf("--------------------------------------------------\n");
	p=&a;
	for(i=0;i<6;i++)
	{
		printf("p[%d]=%d\n",i,p[i]);
	}
	return 0;
}

operation result:

linux@ubuntu:~/workspace/array$ ./main.out
*p=3
*p=-1075484268
*p=2
*p=2
*p=134513888
*p=0
--------------------------------------------------
p[0]=3
p[1]=-1075484272
p[2]=2
p[3]=2
p[4]=134513888
p[5]=0

The results can be seen as two addressing mode.

(gdb) x/3d 0xbffff348
0xbffff348:	0	2	134513920

Achieved begins with the following sequential read address number 3, in decimal (D);

2 pointer arithmetic

int *p=&a;
p++;

After performing execution p ++ pp (gdb view the contents of the pointer p), find the address plus 4;
this is because p is a pointer to the int type, int four bytes each.

p+=3 p[3]Points to an address, but the latter is not shifted p. ''

Array name itself is a pointer constant, can not be changed.

3. Character string arrays and pointers

linux@ubuntu:~/workspace/cstring$ cat main.c
#include<stdio.h>

int main()
{
	char str[]="hello";//字符数组以\0结束；
	char *str2="world";
	char str3[10];
	printf("input the value \n");
	scanf("%s",str3);
	printf("str is %s\n",str);
	printf("str2 is %s\n",str2);
	printf("str3 is %s\n",str3);
	return 0;	
}

1：

scanf("%s",srtr3);

str3 an array name, scanf no need to add the address character & taken;

2：

scanf("%s",srtr3)

The str3 into str, str can also write to the data;
but if str3 into str2, an error occurs, the following reasons;

3：

(gdb) p str
$1 = "hello"
(gdb) p &str
$3 = (char (*)[6]) 0xbffff346
(gdb) x/6c 0xbffff346
0xbffff346:	104 'h'	101 'e'	108 'l'	108 'l'	111 'o'	0 '\000' //字符前面的数字是ASCII码

(gdb) p str2
$2 = 0x8048630 "world"
(gdb) p &str2
$4 = (char **) 0xbffff338

0x8048630 in blocks, the code block is not possible freely modified, so scanf ( "% s", srtr2) to be wrong;

Not a statement:

char *str2;
*str2="abc";

or:

char *str2;
*str2='a';

. 4:
char STR [] and char str3 [10] in the stack memory (variable), it can be modified;
char * str2 in a code (a constant) can not be modified;

5. The in-depth understanding of the array of characters

End of the string \ 0 will not be displayed;
for loop-by-character output when not constrained \ 0;