Answer / guest

#include <stdio.h>

int main(int argc, char** argv)
{
/* This macro B will expand to its argument, followed by a
printf
command that prints the macro invocation as a literal
string */
#define B(x) x; printf(" B(" #x ")\n");

/* This macro A will expand to a printf command that prints
the
macro invocation, followed by the macro argument itself. */
#define A(x) printf(" A(" #x ")\n"); x;

/* Now we call B on a command to print the text of the
program
up to this point. It will execute the command, and then
cause
itself to be printed. */
B(printf("#include <stdio.h>\n\nint main(int argc, char**
argv)\n{\n/*
This macro B will expand to its argument, followed by a
printf\n
command that prints the macro invocation as a literal
string
*/\n#define B(x) x; printf(\" B(\" #x \")\\n\");\n\n/*
This macro
A will expand to a printf command that prints the\n
macro invocation, followed by the macro argument itself.
*/\n#define A(x)
printf(\" A(\" #x \")\\n\"); x;\n\n/* Now we call B on
a command
to print the text of the program\n up to this point. It
will execute
the command, and then cause\n itself to be printed.
*/\n"))
A(printf("/* Lastly, we call A on a command to print the
remainder
of the program;\n it will cause itself to be printed,
and then
execute the command. */\n}\n"))
/* Lastly, we call A on a command to print the remainder of
the program;
it will cause itself to be printed, and then execute the
command. */
}

print the table 5 in loops

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What are the valid places to have keyword “break”?

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Result of the following program is main() { int i=0; for(i=0;i<20;i++) { switch(i) case 0:i+=5; case 1:i+=2; case 5:i+=5; default i+=4; break;} printf("%d,",i); } } a)0,5,9,13,17 b)5,9,13,17 c)12,17,22 d)16,21 e)syntax error

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What are the types of arrays in c?

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what are the stages of compilation

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What is undefined behavior?

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Write a C/C++ program that connects to a MySQL server and displays the global TIMEZONE.

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