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Closed 10 years ago.
Program to understand sizeof operator:
#include<stdio.h>
#include<conio.h>
#include<string.h>
main()
{
char *mess[]={ //array of pointers
"amol is a good boy",
"robin singh",
"genious boy",
"bitch please"
};
printf("%d",sizeof(mess)); // what does sizeof operator do?
}
Please explain the output of this code.
It is the storage size in bytes of 4 pointers to char.
You have your answer right in your question. It has a size of an array of pointers.
So the size is 4 * size of a pointer. (which is 32 on my system.) Your system might vary.
Related
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Closed 9 years ago.
I have a byte array (dB). I am trying to extract the bytes one by one. Why isn't this code working? Any pointers? Logically am I wrong? Or something wrong with my implementation?
You have byte buffers declared like this:
unsigned char *decodeBuf;
To read a single value from that buffer, at offset i you simply write:
unsigned char b = decodeBuf[i];
try
int main()
{
unsigned char tmp;
tmp = getByte(dB+dOffset); dOffset++;
}
it should work
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Closed 10 years ago.
I have an array of chars "0x55".
What I want to do is convert it to a char which is going to be U (because ASCII 0x55 = U).
So how to do this conversion?
#include <windows.h>
int main()
{
array[] = "0x55"
char test;
**// I want to move the string to that test to be one character which is U**
}
Any suggestions?
I think this is what you are after:
int main(int argc,char**argv)
{
char array[] = "0x55";
int value;
char test;
sscanf(array,"%x",&value);
test = value;
return 0;
}
In C++, I would code it a little differently, but this seems more like a C question.
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Closed 10 years ago.
void main(){
int i;
i=printf("how r u?\n");
i=printf("%d",i);
printf("%d",i);}
The above code gives the result as:
how r u?
91
My question:
How does stores 9 and 1??
From the man page: Upon successful return, these functions return the number of characters printed.... If an output error is encountered, a negative value is returned.
So you are getting 9 and i because printf wrote out 9 and 1 characters respectively.
This is also relevant: Why does printf return a value?
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Closed 12 years ago.
i am learning C programming, i was trying to write a recursive function by using this prototype:
void fact(int *n);
The parameter of this function should be passed by reference. Thanks for your help.
I don't feel to be helpful in giving a complete solution -- this is just to show there is an answer:
void fact(int *n)
{
if (*n > 1)
{
int tmp = *n - 1;
fact(&tmp);
*n *= tmp;
}
}
I would never write a factorial function this way.
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Closed 12 years ago.
int main()
{
int x = 2, y = 6, z = 6;
x = y == z;
printf("%d", x);
}
== has higher precedence than =, and y==z is 1.
I will end the answer there, because this looks like homework.
http://codepad.org/fp4ZYJX5
The output is:
1
Have a look at this, which explains a similar, yet more complex question and will answer yours as well.
Output is 1
If you want more help we can give you