This question already has answers here:
C/C++ unsigned integer overflow
(4 answers)
Closed 4 years ago.
I was trying to get into c programming, but in a question a get stuck, please explain this.
int main()
{
char c = 255;
c=c+10;
printf("%d",c);
return 0;
}
the output it gave is
> 9
kindly explain this to me.
The maximum value of a char is 255.
By adding 10 to that number you get 265.
Because that value is not a suitable value for a char it will do 265 % 256 resulting 9
That's why your result is 9
Related
This question already has answers here:
Printing leading 0's in C
(11 answers)
Closed 2 years ago.
my question is short, if I have the following 2 lines of code:
int var = 01;
printf("%d", var);
the output is : 1
how do I get 01 rather than 1?
Use left padded format string.
Solution
int var = 1;
printf("%02d", var);
This question already has answers here:
What does a colon in a struct declaration mean, such as :1, :7, :16, or :32?
(3 answers)
What does ': number' after a struct field mean? [duplicate]
(2 answers)
Closed 6 years ago.
I'm trying to understand the following code in C:
struct values{
int i:3;
int j:3;
int k:2;
};
int main(){
struct values v = {-6,2,5};
printf("%d %d %d", v.i,v.j,v.k);
}
This code produces the following output:
2 2 1
I'm trying to understand what does it mean the assignment for int values used inside the struct i.e.: int i:3 ?
I know that : is not an operator. So what does it do?
Also, can someone explain how this output is achieved?
Thanks very much!
The numbers specifies the length in bits for each field.
Hence i and j are represented in 3 bits while k in 2 bits.
By the way this is question is clearly a duplicate of this question and there's a very good answer I suggest you to read.
This question already has answers here:
printf string, variable length item
(2 answers)
Closed 7 years ago.
I need to
printf(%?d)
Where '?' is some int. How can I do it?
I'm using pure c.
I've tried to work with const char* array. But there was no result.
See the printf man page:
int width = 16;
int value = 42;
printf("%*d\n", width, value);
Output:
42
LIVE DEMO
This question already has answers here:
Printing array elements
(6 answers)
Closed 9 years ago.
The expected output of the following C program is to print the elements in the array. But when actually run, it doesn't do so.
#include<stdio.h>
#define TOTAL_ELEMENTS (sizeof(array) / sizeof(array[0]))
int array[] = {23,34,12,17,204,99,16};
int main()
{
int d;
for(d=-1;d <= (TOTAL_ELEMENTS-2);d++)
printf("%d\n",array[d+1]);//printing the array
return 0;
}//looks simple but no result
What's going wrong? Why am I not getting any output?
In the comparison
d <= (TOTAL_ELEMENTS-2)
TOTAL_ELEMENTS has type size_t so d is converted to unsigned. For, say, sizeof(size_t)==4, this makes the test
0xffffffff < 5
which fails, causing the loop to exit.
If you really want to start your loop counter from -1
d <= (int)(TOTAL_ELEMENTS-2)
would work
This question already has answers here:
Is uninitialized data behavior well specified?
(7 answers)
What will be the value of uninitialized variable? [duplicate]
(6 answers)
Closed 8 years ago.
Please explain why I am getting output 2 here. My expected o/p is 5 or 7. Please throw some light. Thank you!
#include<stdio.h>
typedef enum {a=3, b, c, d, j}e;
void f(e *e1) {
printf("%ld", (int)*e1);
}
main(){
e es;
f(&es);
}
You haven't initialized es, so your program is just printing the random value that happens to be on the stack when the program runs.
You need to say something like:
e es = c;
That will give you the 5 output you seek.