[Solved] Junk values in char* variable

Question

Strings in C need to be NUL terminated. This means you need to add a zero value byte to the end of the string to indicate the end of the string. Because you have no indication of the end of the string when you view/print the value you are reading on past the end of your array into whatever memory is after it.

If the source data contains a NUL terminator you can simply allocate and copy 1 more byte, but assuming it is a fixed length field with no NUL termination you will need to manually add one:

data = new char[lengthOfParam+1];

memcpy(data, &buffer[offset], lengthOfParam);
data[lengthOfParam] = 0;

Also further more looking at this line you posted:

obj[1] = data;

I maybe wrong here and sorry if I am but I strongly suspect this line is not doing what you think it is. This will store a pointer to your string in obj[1] not copy the data from your string. Hence if you delete data, obj[1] would no longer be valid either.

Accepted Answer

Strings in C need to be NUL terminated. This means you need to add a zero value byte to the end of the string to indicate the end of the string. Because you have no indication of the end of the string when you view/print the value you are reading on past the end of your array into whatever memory is after it.

If the source data contains a NUL terminator you can simply allocate and copy 1 more byte, but assuming it is a fixed length field with no NUL termination you will need to manually add one:

data = new char[lengthOfParam+1];

memcpy(data, &buffer[offset], lengthOfParam);
data[lengthOfParam] = 0;

Also further more looking at this line you posted:

obj[1] = data;

I maybe wrong here and sorry if I am but I strongly suspect this line is not doing what you think it is. This will store a pointer to your string in obj[1] not copy the data from your string. Hence if you delete data, obj[1] would no longer be valid either.