数组字符串转换 [英] conversion of array to string

问题描述

 `＃包括LT＆;＆reg51.h GT;
＃包括LT＆;＆string.h中GT;
＃包括_LCD_R8C.c
unsigned char型C [11];
无效serial_int（无效）中断4
{
静态无符号字符CHR [11];
INT I，J;
如果（RI == 1）
{
    RI = 0;
        TI = 0;
    CHR [11] = SBUF;
        为（J = 0; J＆所述; 1; J ++）
         {
       对于（i = 0; I＆LT; = 10;我++）
        {
      C [i] = CHR;
    }
       C [11] ='\\ 0';
     }
   }
}
 

  INT的main（）
{
unsigned char型A [2] [11] = {$○○一六二二一八二六，$ 0123456789};
INT I，J;
    lcd_init（）;
    lcd_clear（）;
    SCON =为0x50;
    TMOD = 0x20的;
    TH1 =是0xFD;
    ET0 = 0;
    TR1 = 1;
    RI = 1;
    ES = 1;
    EA = 1;
    对于（i = 0; I＆LT; = 1;我++）
{
  ** J = STRCMP（A，C）; **
    如果（J == 0）
      {
      lcd_printxy（1,1，是）;
      }
    其他
      {
      lcd_printxy（1,6，否）;
      }
}
}`
 

编译器给警告在第55行（BOLD）：指向不同的对象。
怎么样数组字符串的转换？这是对的吗？
接收到的字符串应该用绳子可用阵列比较..

解决方案

创建一个包含字符串数组，像

 字符富[] =酒吧;
 

它实际上是一样以下

 字符富[] = {'B'，'A'，'R'，'\\ 0'};
 

或以下

 为const char *富=酒吧;
 

这个变量，富可以使用，也可以作为一个数组（即第二个字母是 foo的[1] ），或者可以传递给函数的指针（如的strlen（富）将返回3）。

因此​​，从程序的角度来看，一个字符数组或字符指针是等价的，与你不能分配给一个字符数组，但你可以分配给指针异常。所以以下是好的：

 字符富[] =酒吧;
字符* PFOO = foo的;
 

但是，这是不行的：

 为const char * PFOO =酒吧;
字符富[] = PFOO; / *错误！ * /
 

还有另外一件事那就是字符数组和字符指针之间的不同：一个字符数组的使用的sizeof 运算符返回数组中的字符数（包括终止零），但在一个字符指针它返回指针的大小

例如：

 字符富[] =富;
为const char *巴=酒吧;的printf（的sizeof（富）=％lu个\\ N的sizeof（富））;
的printf（strlen的（富）=％lu个\\ N，strlen的（富））;
的printf（的sizeof（巴）=％lu个\\ N的sizeof（巴））;
的printf（strlen的（巴）=％lu个\\ N的strlen（巴））;
 

如果您运行在64位机器上上述code（其中三分球是八个字节），它将打印：

的sizeof（富）= 4
strlen的（富）= 3
的sizeof（巴）= 8
strlen的（巴）= 3

`#include<reg51.h>
#include<string.h>
#include"_LCD_R8C.c"
unsigned char c[11];
void serial_int (void) interrupt 4
{
static unsigned char chr[11];
int i,j;
if (RI==1)      
{
    RI = 0;     
        TI = 0;     
    chr[11] = SBUF;       
        for(j=0;j<1;j++)
         {
       for(i=0;i<=10;i++)
        {
      c[i]=chr;
    }
       c[11]='\0';
     }
   }
}

int main()
{
unsigned char a[2][11]={"$0016221826","$0123456789"};
int i,j;
    lcd_init();
    lcd_clear();
    SCON = 0x50;              
    TMOD = 0x20;                
    TH1  = 0xFD;                 
    ET0  = 0;                     
    TR1  = 1;                       
    RI   = 1;                   
    ES   = 1;                   
    EA   = 1;                   
    for(i=0;i<=1;i++)
{
  **j=strcmp(a,c);**
    if(j==0)
      {
      lcd_printxy(1,1,"yes");
      }
    else
      {
      lcd_printxy(1,6,"no");
      }
}
}`

the compiler is giving warning in line 55(BOLD): pointer to different objects. how about array to string conversion? is it correct? the received string should compare with the available array of strings..

解决方案

Creating an array that contains a string, like

char foo[] = "bar";

it's actually the same as the following

char foo[] = { 'b', 'a', 'r', '\0' };

or the following

const char *foo = "bar";

This variable, foo can be used either as an array (i.e. the second letter is foo[1]) or as a pointer that can be passed to a function (like strlen(foo) will return 3).

So from the point of your program, a character array or a character pointer are equivalent, with the exception that you can't assign to a character array but you can assign to the pointer. So the following is okay:

char foo[] = "bar";
char *pfoo = foo;

But this is not okay:

const char *pfoo = "bar";
char foo[] = pfoo;    /* Error! */

There is also another thing that is different between character arrays and character pointers: Using the sizeof operator on a character array returns the number of characters in the array (including the terminating zero), but on a character pointer it returns the size of the pointer.

Example:

char foo[] = "foo";
const char *bar = "bar";

printf("sizeof(foo) = %lu\n", sizeof(foo));
printf("strlen(foo) = %lu\n", strlen(foo));
printf("sizeof(bar) = %lu\n", sizeof(bar));
printf("strlen(bar) = %lu\n", strlen(bar));

If you run the above code on a 64-bit machine (where pointers are eight bytes), it will print:

sizeof(foo) = 4
strlen(foo) = 3
sizeof(bar) = 8
strlen(bar) = 3

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