Determine Output:
main()
{
char *str1 = "abcd";
char str2[] = "abcd";
printf("%d %d %d", sizeof(str1), sizeof(str2), sizeof("abcd"));
}
main()
{
char *str1 = "abcd";
char str2[] = "abcd";
printf("%d %d %d", sizeof(str1), sizeof(str2), sizeof("abcd"));
}A. 2 5 5
B. 2 4 4
C. 8 5 5
D. 2 4 5
Answer: Option C
Solution(By Examveda Team)
In first sizeof, str1 is a character pointer so it gives you the size of the pointer variable. In second sizeof the name str2 indicates the name of the array whose size is 5 (including the 'null' termination character). The third sizeof is similar to the second one.
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The answer is 2 5 5
Can you explain how it 8 5 5
2 5 5 is the correct answer
They are using 64 bit compiler so the size of pointer variable is 8 byte in 64 bit compiler... Is it okay friends??
the answer shoud be 2,5,5 as i run the code using tubo c compiler.
cant understand
Hello Shaik,
sizeof(char *) is the size of the pointer, so normally 4 for 32-bit machine, and 8 for 64-bit machine.
its answer is 8 5 5 just know I compile this program