C 中 _int8 数据类型的格式说明符

Question

What is the format specifier for an _int8 data type?

I am using "%hd" but it is giving me an error about stack corruption. Thanks :)

This is a snippet of the code:

signed _int8 answer;

printf("----Technology Quiz----\n\n");
printf("The IPad came out in which year?\n");
printf("Year: ");
scanf("%hd", &answer);
printf("\n\n");
printf("The answer you provided was: %hd\n\n", answer);

Answer 1

man scanf: %hhd "... but the next pointer is a pointer to a signed char or unsigned char". An _int8 is equivalent to a signed char in any system you're going to be doing scanf on.

signed _int8 answer;
scanf("%hhd", &answer);
printf("You entered %d\n\n", answer);

Answer 2

To use the "explicit width" typedefs like int8_t and uint_fast16_t portably in C99 in the format strings of printf and scanf , you need to #include <inttypes.h> and then use the string macros PRIi8 and PRIuFAST16 , like so:

#include <stdint.h>   // for the typedefs (redundant, actually)
#include <inttypes.h> // for the macros

int8_t a = 1;
uint_fast16_t b = 2;

printf("A = %" PRIi8 ", B = %" PRIuFAST16 "\n", a, b);

See the manual for a full list, and cross-reference it with the typedefs in <stdint.h> .

Answer 3

%hd will allow you to get a "short int", which is typically 16 bits, not 8 bits as might imagine. If %hhd is not supported, you may not have a good way to do this, other than scanning in as a short and assigning.

Answer 4

scanf with %hd reads a short int , which might be 16 bit on a 32 bit machine. So you are reading into answer and one byte beyond, thus the stack corruption.

Up to C90 there is no type specifier to read an 8 bit int with scanf.

%hhd is only available since C99, see also ANSI C (ISO C90): Can scanf read/accept an unsigned char? .