c string array length

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c string array length

declare

  • 這是有一個 array 有兩個 items 放 char* 的東西
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    const char *a[2];
    a[0] = "blah";
    a[1] = "hmm";

    const char *strings[] = {"one","two","three"};

    length

    Using sizeof()

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    // Careful though, this works with arrays. It won't work with pointers.
    num = sizeof(names) / sizeof(names[0]);
  • The fact that the strings in the examples are of different length does not matter.
  • names is an array of char *, so the total size of the array in bytes is the number of elements in array times the size of each element (i.e. a char *).
  • Each of those pointers could point to a string of any length, or to NULL. Doesn’t matter.
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    #include<stdio.h>

    int main(void)
    {
        char* names1[]={"A", "B", "C"}; // Three elements
        char* names2[]={"A", "", "C"}; // Three elements
        char* names3[]={"", "A", "C", ""}; // Four elements
        char* names4[]={"John", "Paul", "George", "Ringo"}; // Four elements
        char* names5[]={"", "B", NULL, NULL, "E"}; // Five elements

        printf("len 1 = %zu\n",sizeof(names1)/sizeof(names1[0]));
        printf("len 2 = %zu\n",sizeof(names2)/sizeof(names2[0]));
        printf("len 3 = %zu\n",sizeof(names3)/sizeof(names3[0]));
        printf("len 4 = %zu\n",sizeof(names4)/sizeof(names4[0]));
        printf("len 5 = %zu\n",sizeof(names5)/sizeof(names5[0]));
    }

    // len 1 = 3
    // len 2 = 3
    // len 3 = 4
    // len 4 = 4
    // len 5 = 5

    Using pointer arithmetic

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    int arrSize = *(&arr + 1) - arr;
  • (&arr + 1) points to the memory address right after the end of the array.
  • *(&arr + 1) simply casts the above address to an int *.
  • Subtracting the address of the start of the array, from the address of the end of the array,​ gives the length of the array.

reference