Quick question about atoi

[acidic]

Ok, so I have this array of strings that I want to compare to make sure its a number, so I been thinking atoi, but I cannot get it to compile, how would I compare the stuff inside an array of strings to an interger? The bolded and underlined part is my problem

code:

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


int main(int argc, char *x[]){
if(argc !=3){
printf("Must have two arguments.\n");
exit(0);
}
int d,z,j,l,i,t;
int buffer;

for(i=1;i<argc;i++){
for(j=0;j<4;j++){

                if(x[i][j]=='\0'){
                break;
                }

                                buffer=0;

                        for(t=0;t<10;t++){
[U]                 [B]    if(atoi(x[i][j])==0){[/B][/U]
                                                buffer++;
                                                printf("%d",buffer);
                                                printf("\n%d,%d\n",i,j);}
                                        }

                        if(buffer==0){
                                printf("Arguments must be a number.\n");
                                        exit(0);}


}}
z= atoi(x[1]);
d= atoi(x[2]);

l= z * d;

printf("%d\n",l);


return 0;}

 

[Pyrolistical]

I think you want:

x[i]  // a char *

instead of

x[i][j]  // a char

 

[BillLeeLee]

Just a quick glance, but...

char *argv[] (or x in your code) is not a matrix, it's an array of pointers to char strings).

Also, atoi's return value is whatever the number is contained in the string (if it is indeed a number). It returns 0 only if an error occurs.

 

[acidic]

yea, but I need to compare the stuff in x[row][j] to verify its an interger, using x gets the array, I need the letters that the strings is pointed by the array. That is what I don't know to do.

edit:
what would be a way to check to see if x[j] is an interger?

 

[Lews_Therin]

You are misunderstanding both atoi and character strings.

for(j=0;j<4;j++){ ... }

This will limit the size of the numbers to 4 digits, which you probably don't want.

for(t=0;t<10;t++){
  if(atoi(x[i][j])==0){
    buffer++;
    printf("%d",buffer);
    printf("\n%d,%d\n",i,j);
  }
}

First, this code probably doesn't need the loop. If buffer must be increased 10 times for each "0", then just use buffer += 10

Second, you are calling atoi on a character. It will not convert, for example, the character '3' into the number 3. It will, however, convert the character string "3" into 3, the string "12345" into the integer 12345, etc.

 

[Lews_Therin]

what would be a way to check to see if x[j] is an interger?

Check if its value is between 48 (0) and 57 (9). Eg, if (x[j] >= 48 && x[j] <= 57){ ... }

 

[BillLeeLee]

edit: beaten to it since I was busy coding up something.

atoi expects a char* argument so you have to pass it strings, not just chars.

However, using atoi might not be optimal for your program because it can return 0 if an error occurs, and this can conflict with the input number "0", so "a" and "blah" will produce the same return value as 0.

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

int main(int argc, char *argv[])
{
    while(argc-- && argc)
    {
        if(atoi(argv[argc]) >= 0)
        {
            printf("%s\n", argv[argc]);
        }
    }

    return 0;
}

Just for reference - I'm not recommending the use of this code.

 

[Pyrolistical]

char * x[] is an array of string arguements.

IE. If you ran your programs with the argument foo bar:

c:\> YOUR-PROGRAM foo bar

then

x[0] == "YOUR-PROGRAM"
x[1] == "foo"
x[2] == "bar"

 

[BillLeeLee]

Here's something that'll work with accessing each element of the argv strings. Incredibly hacky, will need to be modified to fit with your code.

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

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

    while(--argc && argc)
    {
        for(i = 0; argv[argc][i] != '\0'; ++i)
        {
            c = argv[argc][i];
            if(c < 48 || c > 57)
            {
                printf("%s is not a number!\n", argv[argc]);
                break;
            }
        }
    }

    return 0;
}

And note that this does not work for negative numbers, or numbers prefixed with the positive sign either, or hex numbers, octal...etc.

 

[acidic]

nvm I found isdigit() function can do it, thanks