通过函数c传递指针

Question

I'm trying to exit a while loop within a function but I don't think i'm writing the pointer correctly, c is new to me and reading examples online is confusing.

I would appreciate feed back on what I'm writing incorrectly.

int logOn(int *log);

int main(void)
{
    int log;
    log = 1;
    while(log = 1)
    {
        int logOn(log);
    }

    while(log = 2)
    {
        int mainMenu();
    }

    printf("\ngood bye");
}

int logOn(*par)
{
    char p;

    printf("would you like to log on (y/n): ");
    scanf("%s", &p);

    if(p="y")
    {
        par = 2;
    }
    else if(p="n");
    {
        par = 3;
    }
    else
    {
        printf("\nError entering command\n")
    }
    return 0;
}

I've updated my code and fixed a lot of the errors you guys helped my identify.

now for some reason it goes through my logon loop twice if i hit a key neither 'y' or 'n'

this is what my current code looks like

include

int logOn(int *log);
int mainMenu(int *log);

int main(void)
{
    int log;
    log = 1;
    while(log == 1)
    {
        (void) logOn(&log);
    }

    while(log == 2)
    {
        mainMenu(&log);
    }

    printf("\ngood bye");
}

int logOn(int *par)
{
    char p;

    printf("would you like to log on (y/n): ");
    scanf("%c", &p);

    if(p == 'y')
    {
        *par = 2;
    }
    else if(p == 'n')
    {
        *par = 3;
    }
    else
    {
        printf("\nError entering command\n");
    }
    return 1;
}

int mainMenu(int *par)
{
    printf("\nMain Menu\n");
    *par = 3;
    return 0;
}

and this is what it is out putting when i dont hit 'y' or 'n'

 would you like to log on (y/n): e

 Error entering command
 would you like to log on (y/n): 
 Error entering command
 would you like to log on (y/n): 

Answer 1

while(log == 1) it's == not = .

When in a while condition you write log = 1 it simply assigns 1 to log . Then while gets 1 from the variable log , which is implicitly converted to true , which makes while loop infinitely.

Answer 2

The parameters of your logOn function must be defined as so:

int logOn(int *par){//function body}

Also the syntax of your while condition must be as follows:

while(log == 1)

Note the double == sign. This is necessary as a single = sign will result in an infinite loop.

Furthermore when you pass in the address of a variable into a function as you have, then a * operator is required in front of it within the function body in order to deference the value at that address.

eg

int logOn(int *par)
{
    printf("%p", par); //address par 
    printf("%d",*par); //value stored at address par
}

Hope this helps.

Answer 3

while(log = 1)
while(log = 2)

Wrong. In c, the = operator is for assignment, not comparison. You are setting log equal to 1 and 2 there. You want the == operator for comparison:

while(log == 1)

Next:

scanf("%s", &p);

Wrong. %s is for arrays of char ending in zero ('\\0'). This is how c represents strings. For a single character, you want

scanf("%c", &p);

Next:

if(p="y")

You are misusing the assignment operator again. In addition, you are comparing a p , a char to "y" , a char * - char array. Double quotes are for strings, single quotes/apostrophes are for single characters. You are looking for

if(p == 'y')

Next:

else if(p="n");

Same errors as above, plus, the semicolon - ; doesn't belong there. Suffice it to say, that should cause a compilation error on your else statement below. You want:

else if(p == 'n')

Next:

par = 2;
par = 3;

You are assigning integers to a pointer. You probably mean

*par = 2;
*par = 3;

Next:

int logOn(log);

should be

(void) logOn(&log);

Next:

int mainMenu();

I don't know what mainMenu is, but the int keyword makes no sense there.

Answer 4

Some points in your program .

First, while(log = 1) , since your are checking if the log is equal to 1 or not, the syntax should be while(log == 1) { }

Secondly, while calling the logOn function, pass the address of log variable ( int logOn(&log) ;)

Third, The signature of the int logOn(*par) function should be int logOn(int *par) .

Fourth, "y" or "n" represents a string literal, if you want to use only a single character, use 'y' or 'n', that is with single quote.

Fifth, To change the value of a pointer use the de-referencing operator * . Your line in the code should be

*par = 2;  (//Content of par equal to 2)

*par = 3;

Happy programming.