如何用C语言编写具有数值和字符值的数组
需要在数组中存储数值和字符值的组合。
我需要将下面的数组存储在atmega328p EEPROM中，但它不是working.The错误状态“多字符字符常量”。col530= {'17'，'12A'，'74'，'23'，'30'，'21'，'31'，'10'，'15'，'33'，'1'，'14'，'11'，'34'，'3'，'9'，'9A'，'11A'，'4A'，‘21’，'16'，'5'，'4AX'，'13'，'75'，'4'，'7'，'6'，'35A'，'8'，'\0'}；
是否有一种将上述值存储在数组中的正确方法？
#include <avr/io.h>
#include <avr/eeprom.h>
// macro for easier usage
#define read_eeprom_array(address,value_p,length) eeprom_read_block ((void *)value_p, (const void *)address, length)
#define write_eeprom_array(address,value_p,length) eeprom_write_block ((const void *)value_p, (void *)address, length)
//declare an eeprom array
float EEMEM SWGA[68];
float EEMEM MA[68];
float EEMEM SWGT[41];
float EEMEM TPCS[41];
float EEMEM Core_area[30];
//char EEMEM *EI_LAM[31];
// declare a ram array and initialize
float col1[28]={47,46,45,44,43,42,41,40,39,38,37,
36,35,34,33,32,31,30,29,28,27,26,25,24,23,22,21,20};
float col2[28]={8,10,13,16,19,24,30,36,39,50,72,90,110,132,151,182,
221,240,290,344,422,508,628,760,940,1210,1600,2030};
float col4[28] = {27546,20223,14392,11457,9337,7755,6543,5595,4838,3507,2800,
2286,1902,1608,1308,1137,997,881,711,609,
504,415,341,286,242,176,137,106};
float col6[30]= {1.213,1.897,2.284,2.723,3.000,3.329,3.703,4.439,4.839,5.880,6.555,6.555,7.259,7.259,7.562,7.865,7.865,9.072,10.284,10.891,10.891,12.704,13.039,14.117,15.324,15.865,18.969,19.356,39.316,40.803};
unsigned char col5[30]= {'17','12A','74','23','30','21','31','10','15','33','1',
'14','11','34','3','9','9A','11A','4A','2','16','5',
'4AX','13','75','4','7','6','35A','8','\0'};
//unsigned char col5[30]= {'017','12A','074','023','030','021','031','010','015','033','001','014','011','034','003','009','09A','11A','04A','002','016','005','4AX','013','075','004','007','006','35A','008','\0'};
// declare another ram array
//float my_other_ram_array[68];
int main(void)
{
// Copy data from my_ram_array to eeprom array
write_eeprom_array(SWGA,col1,sizeof(SWGA));
write_eeprom_array(MA,col2,sizeof(MA));
write_eeprom_array(TPCS,col4,sizeof(TPCS));
write_eeprom_array(Core_area,col6,sizeof(Core_area));
//write_eeprom_array(EI_LAM,col5,sizeof(EI_LAM));
}回答 2
Stack Overflow用户
发布于 2022-05-12 17:54:33
一种方法是使用字符串数组：
char col5[30][4] = {
"17", "12A", "74", "23", "30", "21", "31", "10", "15", "33",
"1", "14", "11", "34", "3", "9", "9A", "11A", "4A", "2", "16",
"5", "4AX", "13", "75", "4", "7", "6", "35A", "8"
};C中的字符串只是一个以零结尾的字符数组.由于您的值都不超过3个字符，长度为4的数组字符串将容纳每个字符串加上终止空字节。
Stack Overflow用户
发布于 2022-05-12 18:12:09
在字符和字符串之间有一种混淆。字符是一个基本的数据类型，大小为1字节，其中字符串是一个以空结尾的字符数组。
您正在尝试编写'17‘，这是非法的，因为char数据类型应该在双引号中只有一个字符，
对于您的情况，您应该选择一个二维字符数组或一个一维字符指针数组。
char array[][4] = {
"17", "12A", "74", "23", "30", "21", "31", "10", "15", "33",
"1", "14", "11", "34", "3", "9", "9A", "11A", "4A", "2", "16",
"5", "4AX", "13", "75", "4", "7", "6", "35A", "8"}；确保列的大小必须大于最大字符串长度，否则会出现溢出
char* array[]={
"17", "12A", "74", "23", "30", "21", "31", "10", "15", "33",
"1", "14", "11", "34", "3", "9", "9A", "11A", "4A", "2", "16",
"5", "4AX", "13", "75", "4", "7", "6", "35A", "8"}；
最好使用指针数组表示法。
https://stackoverflow.com/questions/72219959
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