问题
Just wondering what happens when I use the wrong format specifier in C?
For example:
x = \'A\';
printf(\"%c\\n\", x);
printf(\"%d\\n\", x);
x = 65;
printf(\"%c\\n\", x);
printf(\"%d\\n\", x);
x = 128;
printf(\"%d\\n\", x);
回答1:
what happens when I use the wrong format specifier in C?
Generally speaking, undefined behaviour.*
However, recall that printf is a variadic function, and that the arguments to variadic functions undergo the default argument promotions. So for instance, a char is promoted to an int. So in practice, these will both give the same results:
char x = 'A';
printf("%c\n", x);
int y = 'A';
printf("%c\n", y);
whereas this is undefined behaviour:
long z = 'A';
printf("%c\n", z);
* See for example section 7.19.6.1 p9 of the C99 standard:
If any argument is not the correct type for the corresponding conversion specification, the behavior is undefined.
回答2:
since x is A, the first print f will print: 'A'.
The second will print the ascii value of A (look it up).
The 3rd one will print the ascii character for 65 (I think this is A or a, but its a letter).
The fourth one will print 65.
The 5th one will print 128.
来源：https://stackoverflow.com/questions/16864552/what-happens-when-i-use-the-wrong-format-specifier