Problems in the for loop in language C

Question

The program compiles perfectly, the problem is the cycle, it does not show me the position

here goes the include of library stdio.h

int main(void)
{ int x, i=0,j=0, a[100];
  char sal;

    printf("Este programa lee un arreglo de numeros enteros y un numero x, y muestra en pantalla los indices de las posiciones en donde se encuentra ese numero x\n");
    do
    {
        printf("Ingrese un numero: ");
        scanf("%i", &a[i]);
        i++;
        printf("Si desea salir presione s: ");
        scanf(" %c", &sal);
    }while(sal!='s');   

    printf("Ingrese el valor de la variable x: ");
    scanf("%i", &x);

    for (j; j<=i; j++)
    {
        if(a[i]==x) 
            printf("%i",i);
    }
    printf("\n");
}

It should be for(j; j< i; ++j) because i is the value of the last input plus one. And in the loop it should be if(a[j] == x) printf("%i ", j). Use j, the running index, not i. — Pablo
– Pablo, Commented Feb 19, 2018 at 0:42
regarding: for (j; j<=i; j++) The first parameter j; results in the following output from the compiler: "untitled1.c:18:5: warning: statement with no effect [-Wunused-value" Suggest using: for( ; j<=i; j++ ) — user3629249
– user3629249, Commented Feb 19, 2018 at 2:09
when calling any of the scanf() family of functions: Always check the returned value (not the parameter values) to assure the operation was successful. In all the calls to scanf(), in the posted code, any returned value other than 1 indicates an error occurred — user3629249
– user3629249, Commented Feb 19, 2018 at 2:11
when posting a question about a runtime problem, as this code is doing, Post a minimal reproducible example so we can reproduce the problem.. That includes the #include statements as making us guess as to which header files you actually included can/will result in the question being ignored — user3629249
– user3629249, Commented Feb 19, 2018 at 2:13

Pablo · Accepted Answer · 2018-02-19 01:09:44Z

You condition of the for loop should be j<i, not j<i+1. When the while loop exits, i has the value for the next input, but it has not been set because the loop exited.

Also you are using the index i instead of j in the for-loop. j is the running index, not i:

for (j; j<i; j++)
{
    if(a[j]==x) 
        printf("%i", j);
}

would be correct.

Your while loop is ok, but a bit clunky. First you don't check if you are writing past the limit of a. The conditions should be

while(sal!='s' && i < (sizeof a / sizeof *a));

so that the user cannot input more values than a can hold.

The way you exit the loop is also awkward, the user has to type something different to s to continue and it can only be one character. This would be better:

int c;
char line[100] = { 0 };
do
{
    printf("Ingrese un numero: ");
    scanf("%i", &a[i]);
    while((c = getchar()) != '\n' && c != EOF); // to clear the input buffer
    i++;

    printf("Si desea salir ingrese SALIR. Para continuar presione ENTER: ");
    fgets(line, sizeof line, stdin);

} while(strcmp(line, "SALIR\n") && strcmp(line, "salir\n") && i < (sizeof a / sizeof *a));

Note that strcmp returns 0 when the strings are equal, a non-zero otherwise.

David C. Rankin · Accepted Answer · 2018-02-19 01:00:46Z

1

A tip, you should always set the loop counter to 0 either right before the loop, or in the loop. And it's always a good practice to initialize all variables during declarations. In your case, you did not initialize integer array. Below is a modified version of your code.

int main(void)
{
    int x, i, j;
    x = i = j = 0;

    int a[100] = { 0 };

    char sal = 0;

    printf ("Este programa lee un arreglo de numeros enteros y un numero x, "
            "y muestra en pantalla los indices de las posiciones en donde se "
            "encuentra ese numero x\n");

    while(1)
    {
        printf("Ingrese un numero: ");
        scanf("%d", &a[i]);

        i++;

        printf("Si desea salir presione s: ");
        scanf(" %c", &sal);

        if (sal == 's')
            break;

    }

    printf("Ingrese el valor de la variable x: ");
    scanf("%d", &x);

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

}

edited Feb 19, 2018 at 1:00

David C. Rankin

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answered Feb 19, 2018 at 0:33

Billy

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5 Comments

Matteo Italia Over a year ago

%d and %i are perfectly equivalent in printf; they are both valid in scanf as well, but there %d accepts only decimal values, while %i accepts octal and hexadecimal values as well.

Billy Over a year ago

Hey guys, thanks for information. I have been working in C for past few months, but did not know about %i. Thanks for enlightening me. :)

David C. Rankin Over a year ago

You could handle both 's' and 'S' for exit with if (sal == 's' || sal == 's' - 'a' + 'A')

David C. Rankin Over a year ago

Just another note: always, always check the return of scanf, e.g. if (scanf("%d", &a[i]) != 1) { /* handle error */ } With " %c" you will need to check if not EOF to protect against the user pressing Ctrl+d (or Ctrl+z on windoze) to generate a manual EOF to cancel input.

Billy Over a year ago

Hey David, that is a great suggestion. Error handling is crucial in world of software. Always check return value of methods and handle situations accordingly.

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