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    意法半導體筆試題 簡歷

    時間:2023-04-06 16:45:27 筆試題目 我要投稿
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    意法半導體筆試題 簡歷范文

    A Test for The C Programming Language

    意法半導體筆試題 簡歷范文

      I. History

      1. C was originally designed for and implemented on the (what) operating system on the DEC PDP-11, by (who) .

      2. The most recently approved ANSI/ISO C standard was issued in (when), and single line comments notation “//” is or isn’t a feature of C89.

      II. Syntax and Semantics 1.In a runtime C program, auto variables are stored in , staticvariables are stored in , and function parameters are stored in .

      a. stack b. heap c. neither stack nor heap

      2. The statement “extern int x;” is a , and the keyword extern is used during .

      a. variable declaration b. variable definition c. compilation time d. runtime

      3. There is a complicated declaration: void ( * signal (int, void(*)(int)) ) (int);If a statement “typedef void (*p) (int);” is given,please rewrite this complicated declaration.

      4. The following code is a segment of C program.

      ..........

      void func(int *p)

      {...........}

      ..........

      main()

      {

      int num=0;

      .........

      func(&num);

      ........

      }

      ..........

      Here, the function argument “&num” is passed .

      a. by value b. by reference

      III. Practice

      Create a tree, which has h (h>0) layers, and its each node has w(w>0) sub-nodes.Please complete the following incomplete solution.

      #include

      #include

      struct tree{

      char info;

      p_sub; //link to sub-nodes};

      // allocate memory and initiate

      void dnode ( struct tree* tmp )

      {

      = malloc( sizeof (struct tree) );

      = 0x41;

      = NULL;

      }

      struct tree *dtree (struct tree* subtree, int height, int width)

      {

      int i;

      if ( !subtree ) //if necessary, allocte memory for subtree

      denode(subtree);

      if ( height == 1 )

      return subtree;

      else if ( height == 2 ) {

      struct tree *leaf = NULL;

      for ( i=0; i

      denode ( );

      ;

      leaf = NULL;}

      return subtree;}

      else {

      for ( i=0; i

      }

      return subtree;

      }

      }

      main()

      {

      .........

      struct tree *root = NULL;

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