How to check if argv[count] is an integer

Question

I'm trying to create a command line application in C++ and I want to make sure that the input is an integer after a certain command argument.

For this example, I want to check if the next argument is an integer after the "-p" command argument. Here's the snippet of my code right now.

while (count < argc){
    if (strcmp("-p", argv[count]) == 0){
        has_p = true; //Boolean
        pid = atoi(argv[count + 1]);
        if (pid == 0 && argv[count + 1] != "0" ){
            err = 1;
            cout << "pid argument is not a valid input" << endl;
            pid = -1;
        }
        count++;
    }
...
}

Now this code correctly catches the error in this inputs:

• -p 1777
• -p sss
• -p sss17
• -p [space] -U

but fails at this input format

• -p 17sss

I tried to remedy this by trying to compare it using sprintf. Unfortunately supplying sprintf with the char array pointer only outputs 1 character in buffer2.

while (count < argc){
    if (strcmp("-p", argv[count]) == 0){
        has_p = true; //Boolean
        pid = atoi(argv[count + 1]);
        sprintf(buffer, "%d", pid);
        sprintf(buffer2, "%d", *argv[count + 1]);
        if (pid == 0 && argv[count + 1] != "0" || (buffer != buffer2) ){
            err = 1;
            cout << "pid argument is not a valid input" << endl;
            pid = -1;
        }
        count++;
    }
...
}

Is there a way to make sprintf read the whole char array? If not, is there a better solution for this apart from looping through the pointer until I hit "\0"

Answer 1

atoi() cannot do what you want. You need to use strtol() to achieve this. It has much improved error checking capacity.

Signature:

long int strtol(const char *nptr, char **endptr, int base);

This function returns the converted integral number as a long int value, else zero value is returned. After the conversion, you can check the contents of endptr and decide upon that.

Answer 2

The following is a convenient one-liner...

sscanf(argv[count + 1], "%d%*c", &n) == 1

...that will evaluate true if it was possible to read into the int n and there was no trailing character afterwards. The '*' in %*c suppresses assignment, so there's no need to pass a pointer to a dummy char variable. See scanf docs here.