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I have used the continue and break statements using the if condition inside a for loop. And the condition is to print the variables only when the loop iteration is less than 5.
#include <stdio.h>
int main()
{
int i;
for (i=0;i<10;i++)
{
if (i<5)
{
continue;
}
if (i>5)
{
break;
}
printf("%d is less than 5\n", i);
}
return 0;
}
Why this program prints only once but it should print 5 times.
For i<5 your continue jumps to the next iteration before the print.
For i>5, i.e. i==6, you'll exit the loop.
Thus only for i==5 the print statement will be executed.
Could you please tell me why the pointer in the if statement of the following program is not assigned?
//Detect POSITIVE or NEGATIVE number using ponter
#include <stdio.h>
int main ()
{
char *Number = "POSITIVE";
int a;
printf("Enter a number: ");
scanf("%d", &a);
if (a<0)
{
Number = "NEGATIVE";
}
printf("The number %d is %s", a, Number);
return 0;
}
Is assignment in a if statement is not allowed in C?
This is a separate question, but also easy.
You can assign a value in an if statement, even in the condition, i.e. if (a = 17) is valid - in most such cases you wanted to write the comparison ==.
But in your case you're assigning a const char * to a char *. This is illegal, as the content of const char * can't be modified, whereas the content of char * could be modified.
The simpelst solution would be to declare const char *Number = "POSITIVE";
I would recommend you have a second look on pointers and const, e.g. at Constant pointers vs. pointer to constants in C and C++.
@Debojit Acharjee How many times and in how many places are you going to ask the same question?
If that's the case then how come the pointer is working in the if-statement in the following program:
#include <stdio.h>
int main ()
{
int a=5;
char* b;
if (a==5)
{
b = "six";
printf("%s\n", b);
}
b = "seven";
printf("%s\n", b);
return 0;
}
In the output the text "six" and "seven" are printed even when the pointer is explicitly defined within the if-statement and then again defined. If according to @Dr. Volker Milbrandt pointer is a constant and can't be changed, then why it is changing in this program?
No @Debojit Acharjee not the pointer is constant, the pointer is pointing to something constant. Please read and understand the difference in the already mentioned article Constant pointers vs. pointer to constants. With const and pointers you have to read from right to left: const pointer on const. There are four possibilities:
char* (non-constant) pointer on (non-constant) char arrayconst char* (non-constant) pointer on constant char arraychar* const constant pointer on (non-constant) char arrayconst char* const constant pointer on constant char array
But which constant pointer are you refereeing to in my program? Because I haven't used the "const" token.
Not constant pointer, pointer to constant. And exactely, that const is missing in your code. The fix was mentioned on May 28, 2023, 9:46 AM.
Both code examples below compile without issues with C++14 in Visual Studio 2022.
In former C++ standards implementations, the check was not that strict as today.
//Detect POSITIVE or NEGATIVE number using ponter
#include <stdio.h>
int main()
{
const char* Number = "POSITIVE"; // here you declare and assign a pointer to const "POSITIVE"
int a;
printf("Enter a number: ");
scanf_s("%d", &a);
if (a < 0)
{
Number = "NEGATIVE"; // here you assign a pointer to const "NEGATIVE"
}
printf("The number %d is %s", a, Number);
return 0;
}
#include <stdio.h>
int main()
{
int a = 5;
const char* b; // here you declare a pointer to a constant char array
if (a == 5)
{
b = "six"; // here you assign a pointer to const "six"
printf("%s\n", b);
}
b = "seven"; // here you assign a pointer to const "seven"
printf("%s\n", b);
return 0;
}
I know that but @Dr. Volker Milbrandt said pointer as constant.
The compiler handels each string in your code, like "POSITIVE", as a pointer to a constant, NULL-terminated array of chars.