Char-array到int

Question

我有一个数组char input[11] = {'0','2','7', '-','1','1','2', ,'0','0','9','5'};

如何将输入[0,1,2]转换为int one = 27，输入[3,4,5,6]至int two = -112并输入[7,8,9,10]至int three = 95？

thx，JNK

Answer 1

您可以使用strncpy()的组合来提取字符范围，并使用atoi()将其转换为整数（或读取this question以获取将字符串转换为int的更多方法）

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;
}

输出

Heading: 27
Pitch:   -112
Roll:    95

http://ideone.com/SUutl

Answer 2

据我了解你的评论，你知道第一个条目宽3位，第二个和第三个宽4位：

// 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);