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Suppose I have a C++ program with the following code:
#include <iostream>
void func() {
//code for doing some stuffs
}
int main() {
int a; //line 8
cin>>a;
if(a==5) {
func();
} //line 12
}
In the main() function, after calling the function func() it runs well. But after finishing executing the codes from func() it returns to line 12. But how can I return to line 8 after executing func()? I mean after executing a function, I want the main() function to run from the beginning.
Put everything in your main() in a while(true) loop then your program will loop round to line 8 after finishing func().
Int main() {
while(1){
int a; //line 8
cin>>a;
if(a==5) {
func();
} //line 12
}
}
Here is a basic way, set the amount of times you want the LOOP to run...
#include<iostream>
void func()
{
}
const int LOOP = 10 //for example
int main()
{
for (int i = 0; i < LOOP; i++)
{
int a;
std::cin >> a;
if (a == 5)
func();
}
return 0;
}
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I was expecting 1 2 3 as output, but when I try to run this code:
#include <iostream>
using namespace std;
int fun(int x){
if (x>0){
return fun(x-1);
cout<<x<<endl;
}
}
int main()
{
int x=3;
fun(x);
return 0;
}
I get this warning:
warning: control reaches end of non-void function
Why doesn't it return the value and call fun(x-1)?
But the below code works perfectly. I get 3 2 1 as output.
#include <iostream>
using namespace std;
int fun(int x){
if (x>0){
cout<<x<<endl;
return fun(x-1);
}
}
int main()
{
int x=3;
fun(x);
return 0;
}
Once a function has return'ed, it can't execute any more code:
if (x>0){
return fun(x-1);
cout<<x<<endl; // <-- NEVER EXECUTED
}
The warning is because your function has a non-void return type, but is not return'ing any value when x is <= 0, thus causing undefined behavior.
Try this instead:
#include <iostream>
using namespace std;
int fun(int x){
if (x>0){
int ret = fun(x-1);
cout << x << endl;
return ret;
}
return 0;
}
int main()
{
fun(3);
return 0;
}
Online Demo
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I am trying to print all the subsets of an array. But not getting the correct output.
#include<iostream>
using namespace std;
int arr[]={1,2,3,4,5};
int n=5;
void print(int a[],int cnt, int idx)
{
if(idx==n)
{
for(int i=0;i<cnt;i++)
{
cout<<a[i]<<" ";
}
cout<<endl;
return;
}
a[cnt]=arr[idx];
print(a,cnt,idx+1);
print(a,cnt+1,idx+1);
}
int main()
{
int a[5]={0};
print(a,0,0);
}
the above code only prints "5"
Please help me to rectify the same.
Replace the array with a vector. As the array in C++ is passed by reference, the print function is accessing the same array, which causes the problem.
#include <iostream>
#include<vector>
using namespace std;
int arr[]={1,2,3,4,5};
int n=5;
void print(vector<int>a,int cnt, int idx)
{
if(idx==n)
{
for(int i=0;i<cnt;i++)
{
cout<<a[i]<<" ";
}
cout<<endl;
return;
}
a[cnt]=arr[idx];
print(a,cnt,idx+1);
print(a,cnt+1,idx+1);
}
int main()
{
vector<int >a(5,0);
print(a,0,0);
}
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Check the following code:
#include<iostream>
using namespace std;
class example
{
public:
int number;
example()
{
cout<<"1";
number = 1;
}
example(int value)
{
cout<<"2";
number = value;
}
int getNumber()
{
cout<<"3";
return number;
}
};
int main()
{
example e;
e = 10;
cout<<e.getNumber();
return 0;
}
What is the output of the above code. Also, I want to know what happens when an object is directly assigned to a value. How will the compiler interpret it?
first you typed
example e;
So first constructor called and 1 printed
example()
{
cout<<"1";
number = 1;
}
output :
1
then you typed :
e=10 its equal to e = example(10); so another constructor called :
example(int value) /// beacause you used example(10)
{
cout<<"2";
number = value;
}
so your output is :
12
and number is 2
Finally in :
cout<<e.getNumber();
3 is couted but in the other hand value is `10`
because number = value your number is 10
So in Finally your output is :
12310
thanx for #StoryTeller for editting explaintions
This question already has answers here:
Valgrind reporting a segment overflow
(5 answers)
Closed 6 years ago.
I wonder what this message implies:
==18151== brk segment overflow in thread #1: can't grow to 0x4a26000
Note that the code runs just fine and the output is correct. Should I just ignore this message? And what does it mean?
I think you can ignore it. I got the message in a new allocation in some code that seemed to work perfectly and I also get the message it in the following code:
#include <vector>
struct Something
{
Something() : a1(0), b1(0) { }
unsigned short a1;
unsigned short b1;
};
const int allocsize = 10000;
struct Tester
{
Tester()
{
for (int u = 0; u < allocsize; ++u)
data.push_back(new Something[519]);
}
~Tester()
{
for (int u = 0; u < allocsize; ++u)
delete[] (data[u]);
}
std::vector<Something*> data;
};
void test()
{
Tester t;
// while (true) {;}
}
int main()
{
test();
return 0;
}
I also noticed that others experience the same issue:
Valgrind reporting a segment overflow
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I am studying for an exam and need your help.
I must write my own console terminal in C++, which must work in this way:
Example:
:>plus 5 7 "hit ENTER"
:>12
:>minus 10 12 "hit ENTER"
:>-2
:>combine Hello World "hit ENTER"
:>HelloWorld
:>run netstat "hit ENTER"
:>runs netstat
:>help
:>plus int1 int2
minus int1 int2
combine string1 string2
run ?????
:>exit
program exits
For main block I think it would be something like this
int main(void) {
string x;
while (true) {
getline(cin, x);
detect_command(x);
}
return 0;
}
The functions would be something like this
void my_plus(int a, int b) {
cout << a + b;
}
void my_minus(int a, int b) {
cout << a - b;
}
void my_combine(string a, string b) {
?????????????;
}
void my_run(?????????) {
???????????;
}
And the finally detect_command
void detect_command(string a) {
const int arr_length = 10;
string commands[arr_length] = { "plus", "minus", "help", "exit" };
for (int i = 0; i < arr_length; i++) {
if (a.compare(0, commands[i].length(), commands[i]) == 0) {
?????????????????????;
}
}
}
????? - means I don`t know what to write.
Help to make this program work.
Thanks.
I'm going to use the minus operation as an example...
Make a structure like so:
struct cmd_struct {
const char *name;
void (*func) (void *data);
};
Since your function parameters aren't the same, you gotta make a structure for each, e.g.:
struct minus_op {
int rhs;
int lhs;
};
And use the cmd_struct as an array, like so:
static cmd_struct commands[] = {
{ "minus", &my_minus },
...
};
my_minus would then be:
void my_minus(void *data) {
struct minus_op *mop = data;
... do the computation and return ...
}
And loop through it to detect the command used:
for (int i = 0; i < sizeof(commands) / sizeof(commands[0]); ++i) {
if (strcmp(commands[i].name, a) == 0) {
... prepare the data ...
commands[i].func(data);
}
}
Side Note: In order to get the function parameters from command line, have a splitter, e.g. a white space. Use a vector for this and pass that vector to detect_command
Do also note: Get rid of the void param used in this example and use a char **argv and int argc like in main(). argv would be the arguments, and argc would be the number of arguments passed to the function. e.g. if you say to the program:
>> minus 5 1
Then argc should be 2 (the 5 and the 1) and argv[0] = "5" and argv[1] = "1".
Now that you know the idea behind it, implementing a more flexible version is left to you.
Call a respective function to handle each word. For example:
enum commands {
PLUS,
MINUS,
HELP,
EXIT
//....
};
int detect_command(string a) {
const int arr_length = 10;
string commands[arr_length] = { "plus", "minus", "help", "exit" };
for (int i = 0; i < arr_length; i++) {
if (a.compare(0, commands[i].length(), commands[i]) == 0)
return i;
}
return -1; //unknow word
}
Give the string to detect_command() the function return the respective integer to enum commands (that's our i value) or -1 if word is unknow. Then you can write a function like this to use and process the value determined by detect_command():
void run_command(int cmd)
{
switch(cmd) {
case PLUS: run_plus(); break;
case MINUS: run_minus(); break;
// and so on to all comamnds available
default: error("unknow command");
}
}
each function run_*() should continues the command parsing according to own rules, i.e, the "plus" command should be follow by one integer, one white-space and then another integer, right? run_plus() must validate it and then compute the result. e.g.:
//pseudo code
void run_plus()
{
//input here is one after "plus" word
//here we must validate our valid input: two digits split by a white-spaces
int x = parse_digit();
check(white-space);
int y = parse_digit();
int r = x + y;
display_result(r);
}
NOTE: I'm not a C++ programmer; I did detect_command() code modification to you get my idea. I even don't know if it will compile in C++ for the mismatch types.