字符数组转换为int

Question

I have an array char input[11] = {'0','2','7', '-','1','1','2', ,'0','0','9','5'};

How do I convert input[0,1,2] to int one = 27 , input[3,4,5,6] to int two = -112 and input[7,8,9,10] to int three = 95 ?

thx, JNK

Answer 1

You can use a combination of strncpy() to extract the character range and atoi() to convert it to an integer (or read this question for more ways to convert a string to an int).

int extract(char *input, int from, int length) {
  char temp[length+1] = { 0 };
  strncpy(temp, input+from, length);
  return atoi(temp);
}

int main() {
  char input[11] = {'0','2','7','-','1','1','2','0','0','9','5'};
  cout << "Heading: " << extract(input, 0, 3) << endl;
  cout << "Pitch:   " << extract(input, 3, 4) << endl;
  cout << "Roll:    " << extract(input, 7, 4) << endl;
}

Outputs

Heading: 27
Pitch:   -112
Roll:    95

http://ideone.com/SUutl

Answer 2

As I understand your comment, you know that the first entry is 3 digits wide, the second and third are 4 digits wide:

// not beautiful but should work:
char buffer[5];
int  one   = 0;
int  two   = 0;
int  three = 0;
// read ONE
memcpy(buffer, input, 3); 
buffer[3] = '\0';
one = atoi(buffer);
// read TWO
input += 3;
memcpy(buffer, input, 4);
buffer[4] = '\0';
two = atoi(buffer);
// read THREE
input += 4;
memcpy(buffer, input, 4);
buffer[4] = '\0';
three = atoi(buffer);