数组字符串转换

本文介绍了数组字符串转换的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

`＃包括LT＆;＆reg51.h GT;＃包括LT＆;＆string.h中GT;＃包括_LCD_R8C.cunsigned 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）= 4strlen的（富）= 3的sizeof（巴）= 8strlen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