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...main( ) { int a[3][3]={{1,2},{3,4},{5,6
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2 3 3 4 but the output is: 2 3 5 6 I am expecting the output as 2 3 3 4 because initially the ptr is pointing to the first row of double dimensional array a[][3]. Therefore (*ptr)[1] of first printf will give us 2, like wise (*ptr)[2] will give us 3. But after the ++ptr line it will start pointing to the second element of the first row of a[][3].
To start solving the first problem, given the declaration int a[][3] = {1, 2, 3, 4, 5, 6};, understand how the elements are arranged in the 2-dimensional array based on the provided initialization values.
第1次循环 i =1 进入里面的循环 j = 0 执行s=s+a [1] [0],因为里面循环的结束条件是 j <= i (i = 1),所以里面的循环继续，s = s+a [1] [1],里面的循环执行完毕 接着执行外面的第2次循环，此时 i = 2，执行里面的循环 s = s+a [2] [0],接着里面的循环继续 s = s+a [2] [1] 不知这样写能看明白不，这是一个嵌套的循环
The correct answer is option 1. Concept: An array is defined as a collection of data items of the same type that are stored in contiguous memory locations. Arrays are derived data types in the C programming language that can store primitive data types such as int, char, double, float, and so on. In C programming, you can create an array of arrays. These arrays are known as multidimensional ...
What is printed by the following ANSI C program? #include int main (int argc, char *argv[]) { int a[3][3][3] = {{1, 2, 3, 4, 5, 6, 7, 8, 9}, {10, 11, 12, 13 ...
What is the printout of the following program? public class Test { public static void main (String [] args) { int [] [] values = { {3, 4, 5, 1 }, {33, 6, 1, 2}}; for ...
Step 1: int a [3] [4] = {1, 2, 3, 4, 4, 3, 2, 1, 7, 8, 9, 0}; The array a [3] [4] is declared as an integer array having the 3 rows and 4 colums dimensions. The base address (also the address of the first element) of array is 65472. For a two-dimensional array like a reference to array has type "pointer to array of 4 ints".
Of the following expressions I A [2] II A [2] [3] III B [1] IV B [2] [3] which will not give compile-time errors if used as left hand sides of assignment statements in a C program (GATE CS 2003)?
#include void main () {int a [3] [3]= { {1,2}, {3,4}, {5,6}},i,j,s=0; for (i=1;i<3;i++) for (j=0;j
a [0] [0]=1，a [0] [1]=2，a [0] [3]=0，a [1] [0]=3，a [1] [1]=4，a [1] [2]=0，a [2] [0]=5，a [2] [1]=6，a [2] [2]=0，需要说明的是，只有在调试环境中debug运行才能保证没有赋值的数组元素取值是0，否则它们的值无法确认 s+=a [i] [j];是个累加，s初始值是0，那么分析一下i和j的取值就知道累加了哪些数组元素了 i循环：i=1 ...