How to handle the errors memory cannot be read or written

[bindiya182003]

Why does these 2 errors occur:

1.Memory could not be read
2.Memory could not be written

If it occurs how to handle it

[weaknessforcats]

Memory cannot be read if the address does not belong to your process space.

Memory cannot be written if the address does not belong to your process space.

Use correct addresses.

[JosAH]

Memory cannot be written if the address does not belong to your process space.

Use correct addresses.

Memory also can't be written if a pointer points straight into rom space; emulators
and debuggers are quite smart when it comes to signalling that.

kind regards,

Jos

[bindiya182003]

Memory also can't be written if a pointer points straight into rom space; emulators
and debuggers are quite smart when it comes to signalling that.

kind regards,

Jos

But even for simple input statement i am getting these error..

For eg:
int num;scanf("%d",num);
For this i am getting the same error what should i do to get rid of this..Please help me....Because i can't do any of the programms

[mattmao]

But even for simple input statement i am getting these error..

For eg:
int num;scanf("%d",num);
For this i am getting the same error what should i do to get rid of this..Please help me....Because i can't do any of the programms

Hello man, keep an eye on the correct syntax for the scanf statement!

That is:

int num;
scanf("%d", &num);
 

Note the "&" symbol before the variable name "num". Your lecture notes/reference book would tell you why.

But if you want to print it out:

printf("%d", num);

[bindiya182003]

Hello man, keep an eye on the correct syntax for the scanf statement!

That is:

int num;
scanf("%d", &num);
 

Note the "&" symbol before the variable name "num". Your lecture notes/reference book would tell you why.

But if you want to print it out:

printf("%d", num);

Yes I know '&' specify the address to where the int should be stored. Thus we specify the address-of operator (more on this when we get to pointers). Therefore,
scanf("%d", &num);
will put an int into num correctly

But at times
scanf("%d",num);
this also work smoothly without any error....Why is that?

[mattmao]

Yes I know '&' specify the address to where the int should be stored. Thus we specify the address-of operator (more on this when we get to pointers). Therefore,
scanf("%d", &num);
will put an int into num correctly

But at times
scanf("%d",num);
this also work smoothly without any error....Why is that?

It would be ALWAYS compiled with no error simply because it doesn't violate any syntax check, but this doesn't mean your code is CORRECT.

scanf("%d",num);

This reads: Find the memory address assigned to the "num" pointer variable and then put the value of an integer taken from standard input into the content starts from that position.

Is this what you want to do? I guess not...

AND because you haven't declare the pointer num, its address in memory is not allocated to your program. That is why you got this Segmentation Fault.

So just use

scanf("%d", &num);

scanf statement takes an ADDRESS, not a VARIABLE.




Plus, this is correct:

int a;
int * num = &a;
 
scanf("%d", num);
printf("%d", *num);

Why? Because num now is an ADDRESS.

[bindiya182003]

It would be ALWAYS compiled with no error simply because it doesn't violate any syntax check, but this doesn't mean your code is CORRECT.

scanf("%d",num);

This reads: Find the memory address assigned to the "num" pointer variable and then put the value of an integer taken from standard input into the content of that position.

Is this what you want to do? I guess not...

So just use

scanf("%d", &num);

Thanks a lot for your reply....I really learnt an important thing in C.....