linux下字符串存储的问题
char A[5];
int b = 1234;

如何保存使A[0] = 34
A[1] = 12
A[2] = 00
A[3] = 00
A[4] = 00

即 A[5] = "3412000000";

我保存后总是“D204000000”，郁闷哪，求救，谢谢各位

[解决办法]
仅供参考

C/C++ code

#include <stdio.h>#include <stdlib.h>#include <conio.h>unsigned int iv,i;unsigned char BCD[5];//定长10位BCD码unsigned char sv[11];void main() {    iv=123456789;//最大10位无符号正整数    //INT2BCD    sprintf(sv,"%010u",iv);    for (i=0;i<10;i+=2) {        BCD[i/2]=(sv[i]<<4)|(sv[i+1]&0x0F);    }    printf("BCD=%02x%02x%02x%02x%02x\n",BCD[0],BCD[1],BCD[2],BCD[3],BCD[4]);    //BCD2INT    for (i=0;i<10;i+=2) {        sv[i]='0'|(BCD[i/2]>>4);        sv[i+1]='0'|(BCD[i/2]&0x0F);    }    sscanf(sv,"%010u",&iv);    printf("iv=%010u\n",iv);    getch();}