代码在c中给出意外输出

Question

The following code snippet gives unexpected output in Turbo C++ compiler:

     char a[]={'a','b','c'};
     printf("%s",a);

Why doesn't this print abc ? In my understanding, strings are implemented as one dimensional character arrays in C.
Secondly, what is the difference between %s and %2s ?

Answer 1

This is because your string is not zero-terminated. This will work:

char a[]={'a','b','c', '\0'};

The %2s specifies the minimum width of the printout. Since you are printing a 3-character string, this will be ignored. If you used %5s , however, your string would be padded on the left with two spaces.

Answer 2

char a[]={'a','b','c'};

一个问题是字符串需要以null结尾：

char a[]={'a','b','c', 0};

Answer 3

Without change the original char-array you can also use

     char a[]={'a','b','c'};
     printf("%.3s",a);
or
     char a[]={'a','b','c'};
     printf("%.*s",sizeof(a),a);
or
     char a[]={'a','b','c'};
     fwrite(a,3,1,stdout);
or
     char a[]={'a','b','c'};
     fwrite(a,sizeof(a),1,stdout);

Answer 4

Because you aren't using a string. To be considered as a string you need the 'null termination': '\\0' or 0 (yes, without quotes).

You can achieve this by two forms of initializations:

char a[] = {'a', 'b', 'c', '\0'};

or using the compiler at your side:

char a[] = "abc";

Answer 5

Whenever we store a string in c programming, we always have one extra character at the end to identify the end of the string.

The Extra character used is the null character '\\0' . In your above program you are missing the null character.

You can define your string as

char a[] = "abc";

to get the desired result.