总结命令行

Question

Pretty straightforward, I am trying to sum all of the integers input in the command line. The sum actually works, if I start the program with " 1 1 1 1 " input, the sum increments by one four times. The problem is that sum is initialized at some really large number (4293283588). Why is that?

int main(int argc, char*argv[])
{
  int a = 0;
  int sum = 0;
  size_t i = 0;

  for (i=0; i<argc; i++)
  {
     a = atoi(argv[i]);
     sum = sum + a;
     printf("%ld\n", sum);
  }

  return 0;
}

Answer 1

argv[0] is perhaps the name of the executable. From the standard:

5.1.2.2.1 Program startup

....

If the value of argc is greater than zero, the string pointed to by argv[0] represents the program name ; argv[0][0] shall be the null character if the program name is not available from the host environment. If the value of argc is greater than one, the strings pointed to by argv[1]
through argv[argc-1] represent the program parameters .

Try

for (i=1; i<argc; i++)

---

Also, as @BLUEPIXY indicated, %ld assumes type long . So either change it to %d , or use long sum .