入ってくるデータを解析する関数があります。私の問題は、strncpy を使用した後、印刷しようとするとガベージが発生することです。malloc を使用して、char 配列を正確なサイズにしようとしています。
コード:
void parse_data(char *unparsed_data)
{
char *temp_str;
char *pos;
char *pos2;
char *key;
char *data;
const char newline = '\n';
int timestamp = 0;
temp_str = (char*)malloc(strlen(unparsed_data));
g_print("\nThe original string is: \n%s\n",unparsed_data);
//Ignore the first two lines
pos = strchr(unparsed_data, newline);
strcpy(temp_str, pos+1);
pos = strchr(temp_str, newline);
strcpy(temp_str, pos+1);
//Split the line in two; The key name and the value
pos = strchr(temp_str, ':'); // ':' divides the name from the value
pos2 = strchr(temp_str, '\n'); //end of the line
key = (char*)malloc((size_t)(pos-temp_str)-1); //allocate enough memory
data = (char*)malloc((size_t)(pos2-pos)-1);
strncpy(key, temp_str, (size_t)(pos-temp_str));
strncpy(data, pos + 2, (size_t)(pos2-pos));
timestamp = atoi(data);
g_print("size of the variable \"key\" = %d or %d\n", (size_t)(pos-temp_str), strlen(key));
g_print("size of the variable \"data\" = %d or %d\n", (size_t)(pos2-pos), strlen(data));
g_print("The key name is %s\n",key);
g_print("The value is %s\n",data);
g_print("End of Parser\n");
}
出力:
The original string is:
NEW_DATAa_PACKET
Local Data Set 16-byte Universal Key
Time Stamp (microsec): 1319639501097446
Frame Number: 0
Version: 3
Angle (deg): 10.228428
size of the variable "key" = 21 or 22
size of the variable "data" = 18 or 21
The key name is Time Stamp (microsec)
The value is 1319639501097446
F32
End of Parser
もう一度実行します。
The original string is:
NEW_DATAa_PACKET
Local Data Set 16-byte Universal Key
Time Stamp (microsec): 1319639501097446
Frame Number: 0
Version: 3
Angle (deg): 10.228428
size of the variable "key" = 21 or 25
size of the variable "data" = 18 or 18
The key name is Time Stamp (microsec)ipe
The value is 1319639501097446
F
End of Parser