如果将字符分配给C中的int变量会怎样？

Question

I am trying to get a integer value from argv[1] .

I want to know what happens if the user inputs a character so that I can avoid it. I tried '\\0' and currently this doesn't work.

int main(int argc, char* argv[]){
    int MAX_SIZE;
    MAX_SIZE=atoi(argv[1]);
while(MAX_SIZE=='\0'){
    printf("plz input in correct format: ");
    scanf("%d", &MAX_SIZE);}

Any help would be appreciated.

Answer 1

Your code is a bit weird but from what I understand you want to check if a character is a number or not, for that you can use the function isdigit() to check if the entered value is a number or not, something like this:

char c='a';
if(isdigit(c))    //if true, i.e. it returns non-zero value
    cout<<"number";
else             // if false, i.e. it returns zero
    cout<<"Char";

I have written C++ code as I am more comfortable in C++ but the function isdigit() works in both C and C++.

Answer 2

# include <stdio.h>
# include <stdlib.h>

int main ( int argc, char *argv[] )
{
   int i, n, a;

   printf("\n   argv[0] = %s\n", argv[0] );
   if ( argc <= 1 )
   {
      printf("\n");
      printf("   argc = %d, no arguments given on command line\n", argc );
      printf("\n");
   }

   for ( i = 1; i < argc; i++ )
   {
      n = sscanf( argv[i], "%d", &a );
      if ( n == 1 )
      {
         printf("   read %d off command line in argv[%d]\n", a, i );
      }
      else
      {
         printf("   sscanf failed, n = %d, argv[%d] = %s\n", n, i, argv[i] );
      }
   }

   return 0;
}