Question

I’m using a char[] of size 4 but when I use memcpy() function it stores 8 characters in it and also the character array length becomes 8. What is happing?

I don’t want to use malloc ok.

char strRoh[4]={'\0'};

and then

memcpy(strRoh,Dump+22,4);

Now tell me whats wrong with this

char strIP[]="hhhhhhhh";
char strRoh[4]={'\0'};
char strTheta[4]={'\0'};
char strTimeStamp[6]={'\0'};
char strNMDump[48]={'\0'};  

is there any problem with decelerations cause when i change there order they strings also change there size now strroh is getting 10 chars

what a hell is going on with this

Answer 1

C strings are 0-terminated. This means that if you want to have a string of length n in C, you need n+1 chars for it:

char hello[5] = "hello";

is not a string, because hello has space for 5 chars, and it doesn’t end with 0.

char hello[6] = "hello";

is a string, and has 6 characters: h, e, l, l, o, 0.

To be able to use string related functions in C, you need the terminating 0.

So, change your code to have:

char strRoh[5]={'\0'};
char strTheta[5]={'\0'};
char strTimeStamp[7]={'\0'};
char strNMDump[49]={'\0'};

Note that in C, when you do:

char hello[] = "hello";

the compiler does the counting for you, and makes hello an array of size 6 (one terminating 0):

printf("%zu\n", sizeof hello);

will print 6.