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#include<stdio.h>
int main (void)
{
int i=257;
int *ptr=&i;
printf("%d%d",*((char*)ptr),*((char*)ptr+1));
return 0;
}
Will the output of the above code implementation defined and the output will vary between little endian and big endian machine?
489
Implementation is the execution or practice of a plan, a method or any design, idea, model, specification, standard or policy for doing something. As such, implementation is the action that must follow any preliminary thinking for something to actually happen.
Undefined Behavior results in unpredicted behavior of the entire program. But in unspecified behavior, the program makes choice at a particular junction and continue as usual like originally function executes.
Implementation-defined behavior is defined by the ISO C Standard in section 3.4.1 as: unspecified behavior where each implementation documents how the choice is made. EXAMPLE An example of implementation-defined behavior is the propagation of the high-order bit when a signed integer is shifted right.
A behavior that is appropriately defined should be clear and concise. It should be observable and measurable. Multiple people should be able to observe and measure the same thing. Try to make your definition as specific as you can. This allows you to help the learner make progress more easily.
Yes. The classic way to detect endianness at run time is the following way:
uint32_t var = 1;
uint8_t *ptr = (uint8_t*)&var;
if(*ptr) puts("Little Endian");
else puts("Big Endian");
Even though 257 => 0x0101, an int is most likely 32 bits in which case on a BE machine it will print 00 and on LE, 11.
98
Yes it will. I always write it this way since I am never sure about operator precedence:
*(((char*)ptr)+1)
And to achieve what you want change your %d to a %c in the format string.
42
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