Fails to pass argument to a Function using scanf()

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I can't pass an argument to a function using scanf().


//ADD Function
void add(int x, int y){
    int result = x + y;
    printf("%d + %d = %d\n", x, y, result);
}


int main(void)
{
    printf("Enter The First Number: ");
    scanf("%d", &x);
    printf("Enter The Second Number: ");
    scanf("%d", &y);

    add(x, y);

    return 0;
}

I'm getting Undeclared Error.

Is there a way to fix that? Is it even possible?

You can't use variable x and y in scanf without declaring it.So you have to first declare it before using it in scanf.

#include<stdio.h>

//ADD Function
void add(int x, int y){
int result = x + y;
printf("%d + %d = %d\n", x, y, result);
}


int main(void)
{
    int x;
    int y;
    printf("Enter The First Number: ");
    scanf("%d", &x);
    printf("Enter The Second Number: ");
    scanf("%d", &y);

    add(x, y);

    return 0;
}

C for Fortran Programmers, In general, the printf function can take a variable number of arguments. The scanf statement is the input counterpart of the printf statement. library function but because C does not support passing arguments "by reference" as in FORTRAN. Type of argument; c: Single character: Reads the next character. If a width different from 1 is specified, the function reads width characters and stores them in the successive locations of the array passed as argument. No null character is appended at the end. char * d: Decimal integer: Number optionally preceded with a + or - sign: int * e, E

You need int x, y; in main() so that there are locally scoped (look this up!) variables allocated on the stack, to pass pointers to scanf().

Input and the scanf function, main () { int operand1 = 2, operand2 = 5, result; As you learned earlier, each format specifier in the format string you pass to If the additional arguments are not present with the scanf function, the results are more serious. If the programmer fails to provide enough arguments, or even if the arguments are in the wrong  In general though, scanf should be avoided, since it has problems dealing with rejected input. It can easily work once and fail 5 times afterwards because the first character it reads is instead of a digit. For this reason I suggest the slightly more complicated sscanf, so you can read an entire line then scan it:

  You can write the code that way, and the program will work.
You must declare the variables X Y so that it knows where to input
----------------------------------------------------------------------------
  //ADD Function
    void add(){
            int x,y;
          printf("Enter The First Number: ");
        scanf("%d", &x);
        printf("Enter The Second Number: ");
        scanf("%d", &y);
        int result = x + y;
        printf("%d + %d = %d\n", x, y, result);
    }


    int main(void)
    {

        add();

        return 0;
    }

Passing pointer to a function in C with example, Passing pointers to functions in C programming with example: Pointers can also be passed as an argument to a function. bonus amount will not reflect in the salary, this is because the change made by the function would be done # include <stdio.h> void salaryhike(int *var, int b) { *var = *var+b; } int main() { int salary=0,  The scanf function takes those additional arguments as the addresses of the memory locations into which it should write the user's input. If the programmer fails to provide enough arguments, or even if the arguments are in the wrong order, scanf takes garbage values as addresses.

Just Enough C/C++ Programming, be read, until each field contains meaningful data, the scanf function will not return. Note that a new way of specifying parameters in the function call has crept in. Let's isolate this new functionality in a small example call to scanf so we can in C programming, because you usually pass scalar variables to functions using a  These arguments are expected to be pointers: to store the result of a scanf operation on a regular variable, its name should be preceded by the reference operator (&) (see example). Return Value On success, the function returns the number of items of the argument list successfully filled.

Programming in C, 2/e, lHl main() 2 int *sum(),*s; clrscr(); s=sum(); printf(“Sum = %d”,*s); return 0; I int sum() l int x,y,z,k; printf(“\nEnter Three Values :”); scanf(“%d éd %d”,&x,&y,sz); k=x +y+z; There are two ways in which we can pass arguments to the function. Any change in the formal argument made does not affect the actual arguments  Exercise 4: Modify the source code from scanf() Eats an Integer so that a floating-point number is requested, input, and displayed. You don’t need to prefix a char array variable with an ampersand in the scanf() function; when using scanf() to read in a string, just specify the string variable name.

Rationale for the ANSI C Programming Language, For performance reasons, UNIX does not reset SIGILL to default handling The latter has an extra argument which refers to the "process ID" affected by the signal. portable implementation in C of library functions such as printf and scanf (see machines that may pass arguments in machine registers rather than using the  #include <stdio.h> int main(void) { int value = 0; int *pvalue = NULL; pvalue = &value; /* Set pointer to refer to value */ printf ("Input an integer: "); scanf(" %d

Comments
  • scanf needs the address of the argument to change the value:
  • scanf("%d",&x);
  • also you need to declare and define x and y
  • Oh sorry the address thing wasn't the case. I just forgot to put it in stackoverflow code. And I already declared them on add() Function. Do i need to create a global variable?
  • I am surprised this compiles the way it is. Even defined in the function they are not visible to main?
  • And also initialize x,y with zero.