I was debugging a C program using gdb and I get the following output when I check the value of a variable. What does it signify?

Question

Code:

int a;
int b;
char test[011];
a = 0x41414141;
b = 0x42424242;

gdb output

(gdb) x/s &a
0x7fffffffde1c: "AAAA@\336\377\377\377\177"
(gdb) x/s &b
0x7fffffffde18: "BBBBAAAA@\336\377\377\377\177"

In the code a is initialised with AAAA and b with BBBB . I need to know the following.

1. Why the location of b has BBBBAAAA instead of BBBB it is supposed to have?
2. What does the @\\336\\377\\377\\377\\177 signify?

Answer 1

Why the location of b has BBBBAAAA instead of BBBB it is supposed to have?

It doesn't. The location of b has 0x42424242 (when interpreted as an int ). But by running x/s &b (as opposed to print b ) you are telling gdb to print a string starting at the location of b , rather than to print the int stored there.

It so happens that the bytes stored at the location of b look like "BBBB" when interpreted as ASCII, and the bytes after that look like "AAAA@" when interpreted as ASCII, and then there are some more bytes that aren't printable characters so gdb prints them as escape codes instead, and then there's a 0 byte (which indicates the end of a string).

What does the @\\336\\377\\377\\377\\177 signify?

@ is the character @. \\336 and \\337 and \\177 are escape codes - the bytes following the @ aren't displayable characters, so gdb prints them as octal escape codes instead (using C syntax).