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If your compiler is GCC you can use following syntax:

int array[1024]={[0 ... 1023]=5};

Check out detailed description: 

#include 
   
    #include 
    
     intmain(){    int i = 0;    int array[100] = { [0 ... 99] = 5};     for (i = 0; i < 100; i++) {        printf("array[%d] = %d\n", i, array[i]);    }           return 0;}
    
   

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Unless that value is 0 (in which case you can omit some part of the initializer and the corresponding elements will be initialized to 0), there's no easy way.

Don't overlook the obvious solution, though:

int myArray[10]={
    5,5,5,5,5,5,5,5,5,5};

Elements with missing values will be initialized to 0:

int myArray[10]={
    1,2};//initialize to 1,2,0,0,0...

So this will initialize all elements to 0:

int myArray[10]={
    0};//all elements 0

In C++, an empty initialization list will also initialize every element to 0:

int myArray[10]={};//all elements 0 in C++

Remember that objects with static storage duration will initialize to 0 if no initializer is specified:

staticint myArray[10];//all elements 0

And that "0" doesn't necessarily mean "all-bits-zero", so using the above is better and more portable than memset(). (Floating point values will be initialized to +0, pointers to null value, etc.)

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