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main() {int a[3][2]={0},(*ptr)[2],i,j; for(i=0;i<2;i++) {ptr=a...
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1、首先你声明了int a[3][2]={0}:a是一个二维数组，并且你初始化了这个数组。 (*ptr)[2]，这是一个指向二维数组的指针。 2、for(i=0;i<2;i++) / / 这个函数是你用来从键盘输入参数的
Here's how the expressions relate to each other: ptr == &arr // int (*)[3] == int (*)[3] *ptr == arr // int [3] == int [3] Unless it is the operand of the sizeof or unary & operators, or is a string literal used to initialize a character array in a declaration, an expression of type "N-element array of T" will be converted, or "decay", to an expression of type "pointer to T" and the value of ...
int matrix[2][2], i, j; for (i=0; i < 2; i++) {int *ptr = matrix[0] + 2*i; for (j=0; j < 2; j++) ptr[j] = (i+1)*(j+1);} hey guys, i got across this in one of my past paper questions, however I do not seem to be able to figure this out. Have tried searching for array in matrix but dont seem to understand. ... #include int main(){ int ...
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@AnT I am expecting the output as "2 3 3 4" because I think initially the ptr is pointing to the first row of double dimensional array "a". 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].Therefore (*ptr)[1] of second line should now give us 3 and likewise ...
main() {int a[3][2]={0},(*ptr)[2],i,j; for(i=0;i<2;i++) {ptr=a+i;scanf("%d",ptr);ptr++;} 我来答
Some other issues with the example code. There were a few issues with the code so I went and tidied it up. Assigning out of bounds in an array is undefined behaviour.. test.c:30:3: warning: array index 3 is past the end of the array (which contains 3 elements) [-Warray-bounds] ptr[3] = ccc; ^ ~ test.c:23:3: note: array 'ptr' declared here int (*ptr[3])();
3 3 3 3 4 4 with gcc.But why is the output for the first printf giving 1 1 1 shouldn't it be 4 4 1? lets say if p=6004 and and ptr would be 6004 and ptr++ would be 6008.then ptr-p should give 4.pls correct me.thanks..
Remember array name can easily decays into pointer to first element in most expressions (read some exceptions where array name not decaying into a pointer to first element? ably answered by @H 2 CO 3). For better understanding, consider my diagrams:
int *ptr = (int*)(&a + 1); // what happen here ? The address of the array is taken, and then 1 is added to it, which produces a pointer pointing sizeof a bytes past the beginning of a. That pointer is then cast to an int*, and that is assigned to ptr. The same could be achieved with. int *ptr = &a[5]; in this case.