solidity call使用
solidity call使用
<address>.call(bytes memory payload) returns (bool, bytes memory)
call 仅发送ETH,不传入data
// SPDX-License-Identifier: MIT
pragma solidity ^0.8.26;
contract SendEther {
// 构造函数,payable使得部署的时候可以转eth进去
constructor() payable {}
function callEther(address payable recipient, uint256 amount)
public
returns (bool)
{
(bool success, ) = recipient.call{value: amount}("");
require(success, "Transfer failed.");
return success;
}
}
1、部署合约时附带 value
2、调用合约,给接收地址recipient
转账
call 传入data,附带发送ETH
在合约中调用其他的合约的方法
// SPDX-License-Identifier: MIT
pragma solidity ^0.8.0;
contract Callee {
uint256 value;
function getValue() public view returns (uint256) {
return value;
}
function setValue(uint256 value_) public payable {
require(msg.value > 0, "msg.value must > 0");
value = value_;
}
// 获取合约地址的eth余额
function getBalance() public view returns (uint256) {
return address(this).balance;
}
}
contract Caller006 {
// 构造函数,payable使得部署的时候可以转eth进去
constructor() payable {}
function callSetValue(address callee, uint256 value)
public
returns (bool)
{
Callee meta = Callee(callee);
// 对函数签名和参数进行编码
bytes memory methodop = abi.encodeWithSignature(
"setValue(uint256)",
value
);
(bool flg, ) = address(meta).call{value: 1 ether}(methodop);
if (!flg) {
revert("call function failed");
}
return flg;
}
// 获取合约地址的eth余额
function getBalance() public view returns (uint256) {
return address(this).balance;
}
}
staticcall调用其他合约的方法
// SPDX-License-Identifier: MIT
pragma solidity ^0.8.0;
contract Callee {
function getData() public pure returns (uint256) {
return 42;
}
}
contract Caller {
function callGetData(address callee) public view returns (uint256 data) {
// call by staticcall
// 对函数签名和参数进行编码
bytes memory methodop = abi.encodeWithSignature("getData()");
(bool flg, bytes memory result) = callee.staticcall(methodop);
if(!flg){
revert("staticcall function failed");
}
data = bytesToUint(result);
return data;
}
function bytesToUint(bytes memory b) public pure returns (uint256) {
uint256 number;
for (uint256 i = 0; i < b.length; i++) {
number = number + uint8(b[i]) * (2**(8 * (b.length - (i + 1))));
}
return number;
}
}
原文地址:https://blog.csdn.net/jc0803kevin/article/details/143433966
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