Predict the output of following source code in C language : int main ( ) { char x = 65; printf(“%c |%d|%f”, x, x, x); return 0; }
A | 65 | 0.000000
printf interprets each argument according to its conversion specifier. Arguments of a variadic function undergo default argument promotion, so the char x (value 65) is passed as an int.
Step 1 — %c prints the int as a character; 65 is the ASCII code of A, so A is printed.
Step 2 — %d prints the same value as a decimal integer, 65.
Step 3 — %f expects a double, but an int was passed. No conversion to 65.0 takes place; this is undefined behaviour in the C standard. On the common x86-64 calling convention a double is read from a floating-point register that was never loaded, which typically holds zero, so 0.000000 is printed. The output is therefore A, 65 and 0.000000.
The value 65.000000 would appear only if x were cast to double, and printing 65 for %c ignores that %c prints the character whose code is 65.
A | 65 | 0.000000 option is correct.
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.................... is a client server application that identifies each host on the internet with a unique user-friendly name.
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(i) Kernel
(ii) Interface
(iii) Nucleus
(iv) Compiler
Codes:
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(i) Machine language
(ii) Assembly language
(iii) High level language
(iv) Natural language
Codes:
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| List I | List II |
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| (c) Oula University Library | (iii) Wolf Walk |
| (d) North Carolina State University | (iv) Find the future |
Codes:
Which part of the QR code represents the encoded data?