当C中的输入为0时，为什么我的getc（）总是返回30？

Question

 int main(void){ char buffer[5] = {0}; int i; FILE *fp = fopen("haha.txt", "r"); if (fp == NULL) { perror("Failed to open file \\"mhaha\\""); return EXIT_FAILURE; } for (i = 0; i < 5; i++) { int rc = getc(fp); if (rc == EOF) { fputs("An error occurred while reading the file.\\n", stderr); return EXIT_FAILURE; } buffer[i] = rc; } fclose(fp); printf("The bytes read were... %x %x %x %x %x\\n", buffer[0], buffer[1], buffer[2], buffer[3], buffer[4]); return EXIT_SUCCESS; } 

I put eight 0s in my haha.txt file, and when I run this code it always gives me :

The bytes read were... 30 30 30 30 30

Can someone tell me why?

Answer 1

because '0' == 0x30

The character '0' is the 0x30 (ascii).

Answer 2

'0' that you have entered in your text file is interpreted as char and not as an int . Now, when you try to print that char in its hex (%x) value, the code just finds the equivalent of that char in hex and prints it.

So, your '0' = 0x30 . You can also try giving 'A' in your text file and printing it as hex, you will be getting '41' because 'A' = 0x41 . For your reference, the AsciiTable .

If you want to print out exactly what you have typed in your text file, then simply change your printf("The bytes read were... %x %x %x %x %x\\n", buffer[0], buffer[1], buffer[2], buffer[3], buffer[4]);

To

printf("The bytes read were... %c %c %c %c %c\n", buffer[0], buffer[1], buffer[2], buffer[3], buffer[4]); 
/* %c will tell the compiler to print the chars as char and not as hex values. */

You might want to read Format Specifiers and MSDN Link .

Answer 3

In your printf you use %x and this print in hex . The char '0' is equal to 48 in decimal so 0x30 in hex .

To print 0 in char you need to use printf with %c