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Editorial Team
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Asked: 2026-06-15T17:32:36+00:00

Code: int main(){ short a=1; // #1 char *p=(char*)&a; *(p)=1; // #2 cout <<

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Code:

int main(){
   short a=1;         // #1
   char *p=(char*)&a;
   *(p)=1;            // #2
   cout << a << endl; // Output: 1
   *(p+1)=2;          // #3
   cout << a << endl; // Output: 513
}

From my understanding, the output should be as shown in the picture below, 257 and then 258.
Is there any reason I got different result when I run the program above ?

enter image description here

Update:
I know this is Undefined behavior, but still, does this mean that the decimal to binary conversion is not done as usual: right to left, but instead is done left to right for example:

binary(a)=1000 0000 | 0000 0000

so *(p)=1; will make binary(a)=1000 0000 | 0000 0000 which is 1 in decimal
and *(p+1)=2; will make binary(a)=1000 0000 | 0100 0000 which is 513
which exactly the output of the program.

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1 Answer

  1. Editorial Team

    What happens here is due to the fact that we have a 2-byte short in a little endian CPU architecture. The standard does not require that the architecture be LE, so in any case this program can generate a number of different results when run on different systems.

    A short here is laid out in memory with the least significant byte (LSB) first:

             Memory addresses ------>
            LSB          MSB

         0000 0000   0000 0000

    p points at the LSB and sets is to 1:

             0000 0001   0000 0000

    The result when interpreted as a short is LSB + 256 * MSB, i.e. 1 + 0 * 256 = 1

    p then points at the MSB (which is on the next memory address) and sets is to 2:

             0000 0001   0000 0010

    Result when interpreted as a short: 1 + 2 * 256 = 513

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