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Pointers and Strings (Answer)

  Practice Questions

What is printed by each piece of code?
  1. 
            char fred[50];
    
            strcpy(fred, "Today");
            if (fred == "Today")
                    printf("Of course!\n");
            else
                    printf("Today is not Today on planet C.\n");
    
    Prints Today is not Today on planet C.

    Remember that == between strings compares pointers, not string contents. Since the two strings must be in different locations, the test is false. The correct way to perform this test is to use the string comparison library function. You use #include <string.h> at the top of your program, then code the test if(strcmp(fred, "Today") == 0).

  2. 
            char fred1[] = "barney", fred2[20] = "barney", *fred3 = "barney";
    
            printf("%d %d %d %d %d %d\n", 
                    strlen(fred1), strlen(fred2), strlen(fred3), 
                    sizeof fred1, sizeof fred2, sizeof fred3);
    

    On my PC I get 6 6 6 7 20 4. The last number may vary on different systems, but the others should be the same. The first three numbers represent the length of the string, which is six characters however the string is stored. The 7 represents the space required to store the string, 6 characters plus one byte for the terminator. The 20 is just the storage size of fred2, as declared. Since fred3 is a pointer, the 4 represents the size of one pointer, not the string itself. This will vary from system to system, but 4 is probably the most common. High-end machines (such as the the Alpha) will show 8, and really ancient PC's will show 2. It may even vary between different compilers on the same hardware.

    Note that sizeof is one of the few things in C which treats an array and a pointer differently.

  3. 
            char sue[] = "susan", *p;
    
            for(p = sue; *p++; )
                    printf("%s", p);
            printf("\n");
    
    Prints usansanann

    The test increments p, so that on the first iteration, p is past the first character, and the loop body prints usan. Each successive iteration prints one less character, producing san, an, n, and finally the empty string. When p points to the character n, the test is true, and leaves p pointing to the terminator. This prints out the empty string on the last iteration. Then the test evaluates to the terminator character, and the loop exits.

  4. 
            char bill[50] = "william";
    
            strcpy(bill + 3, "ey coyote");
            printf("%s\n", bill);
    
    Prints wiley coyote

    The expression bill + 3 yields a pointer to position 3 of bill, which is the second l. Then strcpy puts ey coyote in memory starting at that l. The initial wil remains, leaving wiley coyote.

  5. 
            char *sharon = "sharon";
            
            while(*++sharon) printf("%c", sharon[-1]);
            printf("\n");
    

    Prints sharo

    The test advances the pointer sharon before checking what character it points to, thus seeing, in successive tests, h, a, r, o, n, and the terminator. On the terminator, the loop exits. The body is executed with pointer sharon pointing to each of these, but prints the character in front of where sharon points using the expression sharon[-1]. Hence, the body prints s, h, a, r, and o. It does not print n because it precedes the string terminator, and the test fails and terminates the loop before the body can print n.


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