Check if all characters in an array are integers

Question

I'm working on my first simple C program in Visual Studios 2017. The goal is to get two user inputs which have to be integers.

I go through the array and check if every character is an integer but it still returns "The input is not a valid number". Also after retrying 3 times the console crashes.

"projectExample.exe has triggered a breakpoint. occurred" is the displayed message.

int main()
{
    char userInputM[10];
    char userInputN[10];

    //get the value of M from the user
    printf("Enter a value for M: ");
    fgets(userInputM, sizeof(userInputM), stdin);
    //printf(userInputM);

    //check that the entered value for M is a valid positive number
    const int lenOfInput1 = strlen(userInputM);
    for (int i = 0; lenOfInput1; i++) {

        if (!isdigit(userInputM[i])) {
            printf("The input is not a valid number. Try again");

            printf("Enter a value for M: ");
            fgets(userInputM, sizeof(userInputM), stdin);
            printf(userInputM);
        }
    }

    //check that the entered value for N is a number 

    //convert the user input for M to an int for calculation
    //int factorM = atoi(userInputM);
    //printf("%d", factorM);

    //int result = calculate();
    //int a;
    //scanf("%d", &a);
}

Don't forget that `fgets()` will include the newline, and newline is not a digit. Also, don't forget that recalcitrant users might type some blanks or tabs before and/or after the number. You might want to allow that as valid, but not if there are spaces in the number. Inside your loop, you need to re-measure the length of the input after getting more data — you need a `lenOfInput1 = strlen(userInputM);` after the embedded `fgets()`. The loop condition `for (int i = 0; lenOfInput1; i++)` should probably be `for (int i = 0; i < lenOfInput1; i++)` — or you might need to subtract one for `\n`. — Jonathan Leffler, Sep 27 '17 at 23:41
`size_t len; if (!fgets(userInputM, sizeof(userInputM), stdin)) {/*handle error*/} len = strlen(userInputM); if (len && userInputM[len-1] == '\n') userInputM[--len] = 0;` (don't forget `#include ` and do the same for both inputs) — David C. Rankin, Sep 27 '17 at 23:43

J...S · Answer 1 · 2017-09-28T01:08:38.693

If you need to make sure that the number is an integer, you could read in the input into an int with scanf() and check the return value.

int M, rv=-1;
while(rv!=1)
{
    if(rv==0)
    {
        getchar();
    }
    rv=scanf("%d", &M);
}

scanf() would return the number of variables to which values were assigned successfully which will be 1 if the above scanf() is successful.

If the input was not a number and hence the scanf() returns 0, the entered value, which is not a number, is still left unconsumed in the input buffer and that needs to be consumed. It could be done with a getchar(). Otherwise, scanf() will keep trying to read the same value and rv will be 0 each time.

In this way, you don't need a string buffer or a need to use atoi().

Have a look here as well.

As pointed out in the comments, fgets() will read in the trailing new line (\n) into the string and \n is not a digit.

You could replace the \n with \0 like

str[strlen(str)-1]='\0';

I was trying to avoid using scanf() because Visual Studios complaints that its not safe but I'll change the preprocessor definition and try it. — Diego Duarte, Sep 28 '17 at 01:30

score 0 · Answer 2 · answered Sep 28 '17 at 10:42

int count = 0;
int i = 0;
while (userInputM[i])
{
    if (userInputM[i] >= '0' && userInputM[i] <= '9') //if it's an integer
     {
        ++count;  //do something like creating new array and put the
                  //data inside or in my case I count the # of ints
        ++i;
      }    
    else
        ++i;
}

Check if all characters in an array are integers

2 Answers2