Memset function, memset

Memset function, memset

1 // program example 2 # include <stdio. h> 3 # include <string. h> 4 5 int main (void) 6 {7 int I, j, k; 8 char buffer [] = "Helloworld \ n"; 9 printf ("Buffer before memset: % s ", buffer); // output result Buffer before memset: Helloworld10 memset (buffer, '*', strlen (buffer); 11 printf (" Buffer after memset: % s \ n ", buffer); // output result Buffer after memset: * ********** 12 13 // The ch in this section can be of int or other types, not necessarily char type 14 int array [5] =, 3, 5, 2}; 15 (I = 0; I <5; I ++) 16 printf ("% d", array [I]); // output result 1 4 3 5 217 printf ("\ n"); 18 19 memset (array, 0, 5 * sizeof (int); 20 for (j = 0; j <5; j ++) 21 printf ("% d", array [j]); // The output result is 0 0 0 0 022 printf ("\ n "); 23 24 memset (array, 1, 5 * sizeof (int); // use memset to assign values by byte, the value of each array element is actually 0x01010101, that is, decimal 1684300925 for (k = 0; k <5; k ++) 26 printf ("% d", array [k]); // output result 16843009 16843009 16843009 1684300927 printf ("\ n"); 28 return 0; 29} 30/* 31 * Because memset operates on bytes, in bytes, it assigns a value to the four bytes of memory pointed to by array. Each byte is filled with 1 in number, after being converted to binary, 1 is 0000, 32 * an int element is 4 bytes, starting from 0001, 01010101, 16843009, and. If it is converted to hexadecimal, It is 0 x, which is equal, assign a value of 33 * To an int element. Therefore, it is not advisable to use memset to assign an initial value to a non-struct array! 34 */

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