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Views: 1373 | Replies: 9
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Join Date: Oct 2006
Location: Singapore
Posts: 13
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Rep Power: 2
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Hi guys
need a little help for my homework, any help is greatly appreciated.
:cheesy:
Qn 1. pointers
i have a problem with passing pointers to functions
code is as follows:
i keep gettin 2 errors when trying to compile:
a. expected primary expression before '*' token.
b. expected primary expression before "int".
both errors point to the function call.
any ideas/hints on what i'm doin wrong?
_________________________________________________________
Qn2. error checking
i need to error proof my program.
say i need to retrieve an integer input from user.
i need to make sure that the user inputs an integer or i will prompt him to re-enter value:
yet, i'm unable to find a simple way to error proof the above user's input if it is of float type.
i.e. if user enters 2.3, the variable test will automatically be allocated the value 2.
any ideas how to ensure i can detect if user enters a float value?
thanks for any help in advance.
:cheesy:
need a little help for my homework, any help is greatly appreciated.
:cheesy:
Qn 1. pointers
i have a problem with passing pointers to functions
code is as follows:
//function declaration
void computeProduct(product *pProduct,int *pBest);
//function call
computeProduct(product *pProduct,int *pBest);
//function definition
void computeProduct(product *pProduct,int *pBest)
{
//code
}i keep gettin 2 errors when trying to compile:
a. expected primary expression before '*' token.
b. expected primary expression before "int".
both errors point to the function call.
any ideas/hints on what i'm doin wrong?
_________________________________________________________
Qn2. error checking
i need to error proof my program.
say i need to retrieve an integer input from user.
i need to make sure that the user inputs an integer or i will prompt him to re-enter value:
cout << "Enter integer value:";
cin >> test;
while (cin.fail())
{
cin.clear();
cin.ignore(1000,'n' );
cout << "Invalid entry, re-enter value:"
cin >> test;
}yet, i'm unable to find a simple way to error proof the above user's input if it is of float type.
i.e. if user enters 2.3, the variable test will automatically be allocated the value 2.
any ideas how to ensure i can detect if user enters a float value?
thanks for any help in advance.
:cheesy:
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Join Date: Aug 2005
Location: near St Louis, Missouri, USA
Posts: 10,643
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>>//function call computeProduct(product *pProduct,int *pBest);
Do not include the data type in the function call, just the variable name, like this:
Q2: Error checking is a little more difficult to resolve; there is no simple way to do it. In your case I would get keyboard input as a string and then parse the string for illegal characters. Or get keyboard input one character at a time using get() function and test it for illegal characters.
Do not include the data type in the function call, just the variable name, like this:
computeProduct(pProduct, pBest);Q2: Error checking is a little more difficult to resolve; there is no simple way to do it. In your case I would get keyboard input as a string and then parse the string for illegal characters. Or get keyboard input one character at a time using get() function and test it for illegal characters.
I think it's about time we voted for senators with breasts. After all, we've been voting for boobs long enough. ~Clarie Sargent, Arizona senatorial candidate
Those who are too smart to engage in politics are punished by being governed by those who are dumber. ~Plato
Those who are too smart to engage in politics are punished by being governed by those who are dumber. ~Plato
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Join Date: Aug 2005
Location: near St Louis, Missouri, USA
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Not unless you pass a pointer to the pointer in main(). Something like this thread from yesterday.
Also you must make sure the object in the function will not go out of scope then the function returns to main(). Objects created on the stack do not exist beyond the lifetime of the function in which it was declared, so when the function returns to main the pointer will have an invalid address.
Also you must make sure the object in the function will not go out of scope then the function returns to main(). Objects created on the stack do not exist beyond the lifetime of the function in which it was declared, so when the function returns to main the pointer will have an invalid address.
Last edited by Ancient Dragon : Feb 15th, 2007 at 11:34 am.
I think it's about time we voted for senators with breasts. After all, we've been voting for boobs long enough. ~Clarie Sargent, Arizona senatorial candidate
Those who are too smart to engage in politics are punished by being governed by those who are dumber. ~Plato
Those who are too smart to engage in politics are punished by being governed by those who are dumber. ~Plato
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Join Date: Oct 2006
Location: Singapore
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given this simple example:
in the above case, i want the pointer p to point to a variable k of value 3 in the function. i then pass it back to main and display it.
but after assigning the value of p (which is already 3) to another variable x, the value of x becomes a weird value??
any idea what i'm doin wrong?
void changep(int*&p);
int main()
{
int*p;
int x;
changep (p);
cout << *p;
x=*p;
cout << x;
}
void changep(int*&p)
{
int k=3;
p=&k;
}in the above case, i want the pointer p to point to a variable k of value 3 in the function. i then pass it back to main and display it.
but after assigning the value of p (which is already 3) to another variable x, the value of x becomes a weird value??
any idea what i'm doin wrong?
Last edited by ReDeViL : Feb 16th, 2007 at 10:35 am.
This is because you are assigning p the address of a variable k which is local to the function changep and hence k loses its existence (goes out of scope) as soon as the function exits.
"I don't accept change. I don't deserve to live."
"Working a real job is a win if you're lazy, greedy, or unmotivated. If you're average, you fit right in. And if you're above average, the basic terms of employment and premise of the arrangement is against your interests."
"Working a real job is a win if you're lazy, greedy, or unmotivated. If you're average, you fit right in. And if you're above average, the basic terms of employment and premise of the arrangement is against your interests."
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Join Date: Aug 2005
Location: near St Louis, Missouri, USA
Posts: 10,643
Reputation:
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If you make the variable k static then it will work because static variables are similar to globals except they only have scope within the function in which they are declared.
c Syntax (Toggle Plain Text)
void changep(int*&p); int main() { int*p; int x; changep (p); cout << *p; x=*p; cout << x; } void changep(int*&p) { static int k=3; // <<<<<< Here p=&k; }
Last edited by Ancient Dragon : Feb 16th, 2007 at 11:42 pm.
I think it's about time we voted for senators with breasts. After all, we've been voting for boobs long enough. ~Clarie Sargent, Arizona senatorial candidate
Those who are too smart to engage in politics are punished by being governed by those who are dumber. ~Plato
Those who are too smart to engage in politics are punished by being governed by those who are dumber. ~Plato
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then y is it that when i display *p for the 1st time in the main AFTER the function was called, it still manages to display the value or 3?
it only displays rubbish after i try to change the value......
Undefined behaviour. The "it works in this case, so whats wrong with it" thing is not so good to rely on. Since k in your case is an automatic variable whose scope is local to the function you have defined, its existence is and has to be limited to the function block. Any reference to the variable or its location after the function has exited is not to be relied on.
In general, returning the address of the local variable unless you have explicity requested for it (so that you can explicity free it) is a bad progamming practice, resulting in subtle bugs. Better not rely on such things.
"I don't accept change. I don't deserve to live."
"Working a real job is a win if you're lazy, greedy, or unmotivated. If you're average, you fit right in. And if you're above average, the basic terms of employment and premise of the arrangement is against your interests."
"Working a real job is a win if you're lazy, greedy, or unmotivated. If you're average, you fit right in. And if you're above average, the basic terms of employment and premise of the arrangement is against your interests."
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